{"id":2729,"date":"2021-07-23T16:57:57","date_gmt":"2021-07-23T11:27:57","guid":{"rendered":"http:\/\/ncertsolutions.guru\/?p=2729"},"modified":"2021-07-23T18:03:38","modified_gmt":"2021-07-23T12:33:38","slug":"ncert-solutions-for-class-9-maths-chapter-2-hindi","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-hindi\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium)"},"content":{"rendered":"

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (\u092c\u0939\u0941\u092a\u0926) (Hindi Medium)<\/span><\/h2>\n

These Solutions are part of NCERT Solutions for Class 9 Maths in Hindi Medium<\/a>. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.1\u00a0<\/strong><\/span><\/p>\n

Ex 2.1 Class 9 \u0917\u0923\u093f\u0924 Q1. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0935\u094d\u092f\u0902\u091c\u0915\u094b\u0902 \u092e\u0947\u0902 \u0915\u094c\u0928-\u0915\u094c\u0928 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 \u0939\u0948\u0902 \u00a0\u0914\u0930 \u0915\u094c\u0928-\u0915\u094c\u0928 \u0928\u0939\u0940\u0902 \u0939\u0948\u0902 ? \u0915\u093e\u0930\u0923 \u0915\u0947 \u0938\u093e\u0925 \u0909\u0924\u094d\u0924\u0930 \u0926\u0940\u091c\u093f\u090f :
\n(i) 4x2<\/sup>\u00a0\u2013 3x + 7\u00a0<\/strong>
\n(ii) y2<\/sup>\u00a0+\u00a0\u221a2<\/strong>
\n(iii)3\u221at + t\u221a2<\/strong>
\n(iv) y + \\(\\frac { 2 }{ y }\\)<\/strong>
\n(v) x10<\/sup>\u00a0+ y3<\/sup>\u00a0+ t50<\/sup><\/strong>
\n\u0939\u0932:<\/strong>
\n(i) 4x2<\/sup>\u00a0– 3x + 7
\n\u092f\u0939 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902\u00a0\u092c\u0939\u0941\u092a\u0926 \u0939\u0948 \u0915\u094d\u092f\u094b\u0902\u0915\u093f \u091a\u0930 \u0918\u093e\u0924 \u090f\u0915 \u092a\u094d\u0930\u093e\u0915\u0943\u0924 \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 |<\/p>\n

(ii) y2<\/sup>\u00a0+\u00a0\u221a2
\n\u092f\u0939 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902\u00a0\u092c\u0939\u0941\u092a\u0926 \u0939\u0948 \u0915\u094d\u092f\u094b\u0902\u0915\u093f \u091a\u0930 \u0918\u093e\u0924 \u090f\u0915 \u092a\u094d\u0930\u093e\u0915\u0943\u0924 \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 |<\/p>\n

(iii)3\u221at + t\u221a2
\n\u092f\u0939 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 \u0928\u0939\u0940\u0902 \u0939\u0948 \u0915\u094d\u092f\u094b\u0902\u0915\u093f \u091a\u0930 \u0915\u093e \u0918\u093e\u0924 \u090f\u0915 \u092d\u093f\u0928\u094d\u0928\u093e\u0924\u094d\u092e\u0915 \u0938\u0902\u0916\u094d\u092f\u093e \u0939\u0948 \u0915\u094b\u0908 \u092a\u094d\u0930\u093e\u0915\u0943\u0924 \u0938\u0902\u0916\u094d\u092f\u093e \u0928\u0939\u0940\u0902 \u0939\u0948 |<\/p>\n

(iv) y + \\(\\frac { 2 }{ y }\\)
\n\u092f\u0939 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 \u0928\u0939\u0940\u0902 \u0939\u0948 |<\/p>\n

(v) x10<\/sup>\u00a0+ y3<\/sup>\u00a0+ t\u200b50<\/sup>
\n\u092f\u0939 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 \u0928\u0939\u0940\u0902 \u0939\u0948 | \u092c\u0932\u094d\u0915\u093f \u092f\u0939 \u0924\u0940\u0928 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 \u0939\u0948 |<\/p>\n

Ex 2.1 Class 9 \u0917\u0923\u093f\u0924\u00a0Q2. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u092e\u0947\u0902 x2<\/sup>\u00a0\u0915\u093e \u0917\u0941\u0923\u093e\u0902\u0915 \u0932\u093f\u0916\u093f\u090f |<\/strong>
\n(i) 2 + x2<\/sup>\u00a0+ x\u00a0<\/strong>
\n(ii) 2 \u2013 x2<\/sup>\u00a0+ x3<\/sup><\/strong>
\niii) \\(\\frac { \\pi }{ 2 }\\) x2<\/sup>\u00a0+ x<\/strong>
\n(iv)\u00a0\u221a2x \u22121<\/strong>
\n\u0939\u0932:<\/strong>\u00a0<\/strong>
\n(i) 2 + x2<\/sup>\u00a0+ x
\nx2<\/sup>\u00a0\u0915\u093e \u0917\u0941\u0923\u093e\u0902\u0915 =\u00a01<\/p>\n

(ii) 2 \u2013 x2<\/sup>\u00a0+ x3<\/sup>
\nx2<\/sup>\u00a0\u0915\u093e \u0917\u0941\u0923\u093e\u0902\u0915 =\u00a0\u20131\u00a0\u200b<\/p>\n

(iii) \\(\\frac { \\pi }{ 2 }\\) x2<\/sup>\u00a0+ x
\nx2<\/sup>\u00a0\u0915\u093e \u0917\u0941\u0923\u093e\u0902\u0915 = \\(\\frac { \\pi }{ 2 }\\)<\/p>\n

(iv)\u00a0\u221a2x \u22121
\nx2<\/sup>\u00a0\u0915\u093e \u0917\u0941\u0923\u093e\u0902\u0915 =\u00a00 [\u0915\u094d\u092f\u094b\u0902\u0915\u093f \u092f\u0939\u093e\u0901 x2\u00a0<\/sup>\u0928\u0939\u0940\u0902 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f \u0907\u0938\u0915\u093e \u0917\u0941\u0923\u093e\u0902\u0915 \u0936\u0942\u0928\u094d\u092f \u0939\u094b\u0917\u093e |]<\/p>\n

Ex 2.1 Class 9 \u0917\u0923\u093f\u0924\u00a0Q3. 35 \u0918\u093e\u0924 \u0915\u0947 \u0926\u094d\u0935\u093f\u092a\u0926 \u0915\u093e \u0914\u0930 100 \u0918\u093e\u0924 \u0915\u0947 \u090f\u0915\u092a\u0926\u0940 \u0915\u093e \u090f\u0915-\u090f\u0915 \u0909\u0926\u093e\u0939\u0930\u0923 \u0926\u0940\u091c\u093f\u090f|<\/strong>
\n\u0939\u0932:<\/strong>
\n35 \u0918\u093e\u0924 \u0915\u093e \u090f\u0915 \u0926\u094d\u0935\u093f\u092a\u0926\u0940
\n\u21d2 2x35<\/sup>\u00a0+ 5y\u00a0\u200b
\nNote:<\/strong>\u00a0\u0926\u094d\u0935\u093f\u092a\u0926\u0940 \u0915\u093e \u0905\u0930\u094d\u0925 \u0926\u094b \u092a\u0926\u094b\u0902 \u0935\u093e\u0932\u093e \u0935\u094d\u092f\u0902\u091c\u0915 \u091c\u0948\u0938\u0947 –\u00a0\u00a0x + 5, 3a – 2b, 3t + 7 \u0906\u0926\u093f.
\n100 \u0918\u093e\u0924 \u0915\u093e \u090f\u0915 \u090f\u0915\u092a\u0926\u0940
\n\u21d2\u00a03y100<\/sup>
\nNote:<\/strong>\u00a0\u090f\u0915\u092a\u0926\u0940 \u0915\u093e \u0905\u0930\u094d\u0925 \u090f\u0915 \u092a\u0926 \u0935\u093e\u0932\u093e \u0935\u094d\u092f\u0902\u091c\u0915 \u091c\u0948\u0938\u0947-\u00a03x, 5t, y, 3xy \u0906\u0926\u093f.<\/p>\n

Ex 2.1 Class 9 \u0917\u0923\u093f\u0924\u00a0Q4. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092c\u0939\u0941\u092a\u0926\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0947 \u0918\u093e\u0924 \u0932\u093f\u0916\u093f\u090f:
\n(i) 5x3<\/sup>\u00a0+ 4x2<\/sup>\u00a0+ 7x<\/strong>
\n(ii) 4 \u2013 y2<\/sup><\/strong>
\n(iii) 5t \u2013\u00a0\u221a7<\/strong>
\n(iv) 3<\/strong>
\n\u0939\u0932:<\/strong>
\n(i) 5x3<\/sup>\u00a0+ 4x2<\/sup>\u00a0+ 7x
\n\u0909\u0924\u094d\u0924\u0930: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0918\u093e\u0924 = 3
\n[\u0928\u094b\u091f:<\/strong>\u00a0\u092c\u0939\u0941\u0924 \u0915\u093e \u0918\u093e\u0924 \u091c\u094d\u091e\u093e\u0924 \u0915\u0930\u0928\u0947 \u0915\u0947 \u0932\u093f\u090f \u0938\u092d\u0940 \u0918\u093e\u0924\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u0938\u092c\u0938\u0947 \u092c\u095c\u0940 \u0918\u093e\u0924 \u0915\u094b \u091a\u0941\u0928\u093e \u091c\u093e\u0924\u093e \u0939\u0948 |]<\/p>\n

(ii) 4 \u2013 y\u200b2<\/sup>
\n\u0909\u0924\u094d\u0924\u0930: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0918\u093e\u0924 = 2<\/p>\n

(iii) 5t \u2013\u00a0\u221a7
\n\u0909\u0924\u094d\u0924\u0930: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0918\u093e\u0924 = 1<\/p>\n

(iv) 3
\n\u0909\u0924\u094d\u0924\u0930: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0918\u093e\u0924 = 0
\n[\u0928\u094b\u091f: \u091a\u0942\u0901\u0915\u093f \u092f\u0939\u093e\u0901 \u0915\u094b\u0908 \u091a\u0930 \u0928\u0939\u0940\u0902 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0918\u093e\u0924 \u0936\u0942\u0928\u094d\u092f (0) \u0939\u0948 |]<\/p>\n

Ex 2.1 Class 9 \u0917\u0923\u093f\u0924\u00a0Q5. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u094b \u0930\u0948\u0916\u093f\u0915, \u0926\u094d\u0935\u093f\u0918\u093e\u0924 \u0914\u0930 \u0924\u094d\u0930\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926 \u092e\u0947\u0902 \u0935\u0930\u094d\u0917\u0940\u0915\u0943\u0924 \u0915\u0940\u091c\u093f\u090f:<\/strong>
\n(i) x\u200b2<\/sup>\u00a0+ x\u00a0<\/strong>
\n(ii) x \u2013 x3<\/sup>\u00a0<\/strong>
\n(iii) y + y2<\/sup>\u00a0+ 4<\/strong>
\n(iv) 1 + x<\/strong>
\n(v) 3t<\/strong>
\n(vi) r2<\/sup>\u00a0<\/strong>
\n(vii) 7x2<\/sup><\/strong>
\n\u0939\u0932:<\/strong>
\n(i) x\u200b2<\/sup>\u00a0+ x
\n\u0909\u0924\u094d\u0924\u0930: \u0926\u094d\u0935\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926<\/p>\n

(ii) x \u2013 x3<\/sup>
\n\u0909\u0924\u094d\u0924\u0930: \u0924\u094d\u0930\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926<\/p>\n

(iii) y + y2<\/sup>\u00a0+ 4
\n\u0909\u0924\u094d\u0924\u0930: \u0926\u094d\u0935\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926<\/p>\n

(iv) 1 + x
\n\u0909\u0924\u094d\u0924\u0930: \u0930\u0948\u0916\u093f\u0915 \u092c\u0939\u0941\u092a\u0926<\/p>\n

(v) 3t
\n\u0909\u0924\u094d\u0924\u0930: \u0930\u0948\u0916\u093f\u0915 \u092c\u0939\u0941\u092a\u0926<\/p>\n

(vi) r2<\/sup>
\n\u0909\u0924\u094d\u0924\u0930: \u0926\u094d\u0935\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926<\/p>\n

(vii) 7x2<\/sup>
\n\u0909\u0924\u094d\u0924\u0930: \u0924\u094d\u0930\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.2<\/strong><\/span><\/p>\n

Ex 2.2 Class 9 \u0917\u0923\u093f\u0924 \u092a\u094d\u09301. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092a\u0930 \u092c\u0939\u0941\u092a\u0926 5x \u2013 4x2<\/sup>\u00a0+ 3 \u0915\u0947 \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
\n(i) x = 0
\n(ii) x = \u20131
\n(iii) x = 2<\/strong>
\n\u0939\u0932:<\/strong>
\n(i) p(x) = 5x – 4x2\u00a0<\/sup>+ 3
\n\u092c\u0939\u0941\u092a\u0926 p(x) \u092e\u0947\u0902 x = 0 \u0930\u0916\u0928\u0947 \u092a\u0930
\nP(0) = 5(0) – 4(0)2\u00a0<\/sup>+ 3 = 0 – 0 + 3 = 3
\n\u0905\u0924: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u092e\u093e\u0928 3 \u0939\u0948 |<\/p>\n

(ii) p(x) = 5x – 4x2\u00a0<\/sup>+ 3
\n\u092c\u0939\u0941\u092a\u0926 p(x) \u092e\u0947\u0902 x = -1 \u0930\u0916\u0928\u0947 \u092a\u0930
\nP(1) = 5(-1) – 4(-1)2\u00a0<\/sup>+ 3 = – 5 – 4 + 3 = – 9\u00a0+ 3 = – 6
\n\u0905\u0924: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u092e\u093e\u0928 – 6\u00a0\u0939\u0948 |<\/p>\n

(iii) p(x) = 5x – 4x2\u00a0<\/sup>+ 3
\n\u092c\u0939\u0941\u092a\u0926 p(x) \u092e\u0947\u0902 x = 2 \u0930\u0916\u0928\u0947 \u092a\u0930
\nP(2) = 5(2) – 4(2)2\u00a0<\/sup>+ 3 = 10 -16 + 3 = – 3
\n\u0905\u0924: \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u092e\u093e\u0928 – 3\u00a0\u0939\u0948 |<\/p>\n

Ex 2.2 Class 9 \u0917\u0923\u093f\u0924\u00a0Q2. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092c\u0939\u0941\u092a\u0926\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0947 \u0932\u093f\u090f p(0), p(1) \u0914\u0930 p(2) \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f|\u00a0<\/strong>
\n(i)\u00a0 p(y) = y2<\/sup>\u00a0\u2013 y + 1<\/strong>
\n(ii) p(t) = 2 + t + 2t\u00a02<\/sup>\u00a0\u2013 t\u00a03<\/sup><\/strong>
\n(iii) p(x) = x3<\/sup><\/strong>
\n(iv) p(x) = (x \u2013 1) (x + 1)<\/strong>
\n\u0939\u0932:<\/strong>
\n(i) p(y) = y2\u00a0<\/sup>– y + 1
\nP(0) \u0915\u0947 \u0932\u093f\u090f
\nP(0) = (0)2<\/sup>– 0 + 1 = 1
\nP(1) \u0915\u0947 \u0932\u093f\u090f
\nP(1) = (1)2<\/sup>– 1 + 1
\n= 1 – 1 + 1 = 1
\nP(2) \u0915\u0947 \u0932\u093f\u090f
\nP(2) = (2)2<\/sup>– 2 + 1
\n= 4 – 2 + 1 = 3<\/p>\n

(ii) p(t) = 2 + t + 2t2<\/sup>– t3<\/sup>
\nP(0) \u0915\u0947 \u0932\u093f\u090f
\nP(0) = 2 + 0 + 2(0)2<\/sup>– (0)3<\/sup> = 2
\nP(1) \u0915\u0947 \u0932\u093f\u090f
\nP(1) = 2 + 1 + 2(1)2\u00a0<\/sup>– (1)3<\/sup> = 4<\/p>\n

P(2) \u0915\u0947 \u0932\u093f\u090f
\nP(2) = 2 + 2 + 2(2)2<\/sup>– (2)3<\/sup>
\n= 4 + 8 – 8 = 4<\/p>\n

(iii) p(x) = x3<\/sup>
\nP(0) \u0915\u0947 \u0932\u093f\u090f
\nP(0) = (0)3<\/sup>\u00a0= 0
\nP(1) \u0915\u0947 \u0932\u093f\u090f
\nP(1) = (1)3\u00a0<\/sup>= 1
\nP(2) \u0915\u0947 \u0932\u093f\u090f
\nP(2) = (2)3\u00a0<\/sup>= 8<\/p>\n

(iv) P(x) = (x \u2013 1) (x + 1)
\nP(0) \u0915\u0947 \u0932\u093f\u090f
\nP(0) = (0 – 1) (0 + 1) = (-1) (1) = -1
\nP(1) \u0915\u0947 \u0932\u093f\u090f
\nP(1) = (1 – 1) (1 + 1) = 0 (1) = 0
\nP(2) \u0915\u0947 \u0932\u093f\u090f
\nP(2) = (2 – 1) (2 + 1) = 1(3) = 3<\/p>\n

Ex 2.2 Class 9 \u0917\u0923\u093f\u0924\u00a0Q3. \u0938\u0924\u094d\u092f\u093e\u092a\u093f\u0924 \u0915\u0940\u091c\u093f\u090f \u0915\u093f \u0926\u093f\u0916\u093e\u090f \u0917\u090f \u092e\u093e\u0928 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u094d\u0925\u093f\u0924\u093f\u092f\u094b\u0902 \u092e\u0947\u0902 \u0938\u0902\u0917\u0924 \u092c\u0939\u0941\u092a\u0926 \u0915\u0947 \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948\u0902 :<\/strong>
\n\"NCERT
\n\u0939\u0932: \u00a0<\/strong>
\n(i) P(x) = 3x + 1
\n\"Maths
\np(x) = 0, \u0905\u0924: \u0926\u093f\u092f\u093e \u0917\u092f\u093e x \u0915\u093e \u092e\u093e\u0928 \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n(ii) P(x) = 5x\u00a0–\u00a0\u03c0
\n\"NCERT
\n=\u00a05\u00a0–\u00a0\u03c0
\n\u2235\u00a0\u200bP(x)\u00a0\u2260\u00a00
\n\u2234 x \u0915\u0947 \u0932\u093f\u090f \u0926\u093f\u092f\u093e \u0917\u092f\u093e \u092e\u093e\u0928 P(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0928\u0939\u0940\u0902 \u0939\u0948|<\/p>\n

(iii) P(x) = x2\u00a0<\/sup>– 1
\n\"Class
\n\"NCERT
\n\"NCERT
\n\"NCERT
\n\"NCERT<\/p>\n

Ex 2.2 Class 9 \u0917\u0923\u093f\u0924\u00a0Q4. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u094d\u0925\u093f\u0924\u093f\u092f\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0938\u094d\u0925\u093f\u0924\u093f \u092e\u0947\u0902\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
\n(i) P(x) = x\u00a0+\u00a05<\/strong>
\n(ii) P(x) = x\u00a0\u2013\u00a05<\/strong>
\n(iii) Px) = 2x + 5<\/strong>
\n(iv) P(x) = 3x\u00a0–\u00a02<\/strong>
\n(v) P(x) = 3x<\/strong>
\n(vi) P(x) = ax, a\u00a0\u2260\u00a00<\/strong>
\n\u0939\u0932 (i) :<\/strong>
\n(i)\u00a0\u00a0 P(x) = x + 5
\n\u21d2\u00a0x + 5 = 0
\n\u21d2 x = – 5
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 – 5 \u0939\u0948\u0902 |<\/p>\n

\u00a0\u0939\u0932\u00a0(ii) :<\/strong>
\n(ii) P(x) =\u00a0x\u00a0–\u00a05
\n\u21d2\u00a0x\u00a0–\u00a05 \u00a0= 0
\n\u21d2 x = 5
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915\u00a05 \u0939\u0948|<\/p>\n

\"Maths
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \\(\\frac { -5 }{ 2 }\\) \u0939\u0948 |<\/p>\n

(iv)\u00a0 P(x) = 3x – 2
\n3x – 2 = 0 \u2260
\n\"NCERT
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \\(\\frac { 2 }{ 3 }\\) \u0939\u0948 |
\n\"Maths
\n\"Maths<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.3<\/strong><\/span><\/p>\n

Ex 2.3 Class 9 \u0917\u0923\u093f\u0924 Q1. x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1 \u0915\u094b \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 \u0936\u0947\u0937\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :\u00a0<\/strong>
\n(i) x + 1\u00a0<\/strong>
\n(ii) x –\u00a0\\(\\frac { 1 }{ 2 }\\) <\/strong>
\n(iii) x\u00a0<\/strong>
\n(iv) x +\u00a0\u03b8<\/strong>
\n(v) 5 + 2x\u00a0<\/strong>
\n\u0939\u0932 :\u00a0(i)\u00a0x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1 \u0915\u094b x + 1 \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930\u00a0<\/strong>
\n\"Polynomials
\n\u0905\u0924: \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 \u0936\u0947\u0937\u092b\u0932 0 \u0939\u0948|<\/p>\n

\"Polynomials<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iii)\u00a0\u00a0<\/strong>x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1 \u0915\u094b x \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930\u00a0\u00a0 \u00a0 \u00a0\u00a0<\/strong>
\n\"Class
\n\u0905\u0924: \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 \u0936\u0947\u0937\u092b\u0932 1 \u0939\u0948|<\/p>\n

\u0939\u0932 :\u00a0(iv) x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1 \u0915\u094b\u00a0x +\u00a0<\/strong>\u03c0\u00a0\u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930\u00a0<\/strong>
\n\"NCERT
\n\u0905\u0924: \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 \u0936\u0947\u0937\u092b\u0932\u00a0–\u00a0<\/strong>\u03c03<\/sup>\u00a0+ 3\u03c02<\/sup>\u00a0– 3\u03c0\u00a0+ 1 \u0939\u0948|
\n\u0939\u0932 :<\/strong>\u00a0(v)\u00a0<\/strong>x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1 \u0915\u094b\u00a0<\/strong>5<\/strong>\u00a0+\u00a0<\/strong>2x\u00a0<\/strong>\u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930<\/strong>
\n\"NCERT
\n\"Class
\n\"NCERT<\/p>\n

Ex 2.3 Class 9 \u0917\u0923\u093f\u0924\u00a0Q2.\u00a0<\/strong>x3<\/sup>\u00a0<\/strong>–<\/strong>\u00a0ax2<\/sup>\u00a0+ 6x\u00a0<\/strong>–<\/strong>\u00a0a\u00a0<\/strong>\u00a0\u0915\u094b x\u00a0<\/strong>–<\/strong>\u00a0a \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 \u0936\u0947\u0937\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f |<\/strong>
\n\u0939\u0932 :<\/strong>\u00a0p(x) = x3<\/sup>\u00a0– ax2<\/sup>\u00a0+ 6x – a \u00a0\u0914\u0930 g(x) = x – a \u0939\u0948 |
\ng(x) = x – a \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915
\n\u0905\u0924:\u00a0 x – a = 0
\nx = a
\n\u0905\u0924: \u0936\u0947\u0937\u092b\u0932 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947
\np(x) \u0915\u094b x – a \u0938\u0947 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930 \u0936\u0947\u0937\u092b\u0932 \u092a\u094d\u0930\u092e\u0947\u092f \u0926\u094d\u0935\u093e\u0930\u093e \u0936\u0947\u0937\u092b\u0932 p(a) \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0939\u094b\u0917\u093e |
\n\u0907\u0938\u0932\u093f\u090f, p(a) = (a)3<\/sup>\u00a0– a(a)2<\/sup>\u00a0+ 6(a) – a
\n= a3<\/sup>\u00a0– a3<\/sup>\u00a0+ 6a – a = 5a
\n\u0905\u0924: \u0936\u0947\u0937\u092b\u0932 5a \u0939\u0948 |<\/p>\n

\"Class<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.4<\/strong><\/span><\/p>\n

Ex 2.4 Class 9 \u0917\u0923\u093f\u0924 Q1. \u092c\u0924\u093e\u0907\u090f \u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092c\u0939\u0941\u092a\u0926\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u0915\u093f\u0938 \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 x + 1 \u0939\u0948|<\/strong>
\n(i) x3<\/sup>\u00a0+ x2<\/sup>\u00a0+ x + 1<\/strong>
\n(ii) x4<\/sup>\u00a0+ x3<\/sup>\u00a0+ x2<\/sup>\u00a0+ x + 1<\/strong>
\n(iii) x4<\/sup>\u00a0+ 3x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ x + 1<\/strong>
\n(iv) x3<\/sup>\u00a0– x3<\/sup>\u00a0– (2 +\u00a0\u221a2)x +\u00a0\u221a2<\/strong>
\n\u0939\u0932 :<\/strong>\u00a0(i) p(x) = x3<\/sup>\u00a0+ x2<\/sup>\u00a0+ x + 1
\n\u092e\u093e\u0928\u093e g(x) = x + 1 = 0
\n\u21d2 x = – 1
\n\u0905\u092c \u0917\u0941\u0923\u0928\u0916\u0923\u094d\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947
\np(x) = 0 \u092f\u0926\u093f x = -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0905\u0924: p(x) \u092e\u0947\u0902 x = -1 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(x) = x3<\/sup>\u00a0+ x2<\/sup>\u00a0+ x + 1
\np(-1) = (-1)3<\/sup>\u00a0+ (-1)2<\/sup>\u00a0+ (-1) + 1
\n= – 1 + 1 – 1 + 1 = 0
\n\u091a\u0942\u0901\u0915\u093f p(-1) = 0 \u0907\u0938\u0932\u093f\u090f -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 \u0914\u0930 x + 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(ii) p(x) = x4<\/sup>\u00a0+ x3<\/sup>\u00a0+ x2<\/sup>\u00a0+ x + 1
\n\u092e\u093e\u0928\u093e g(x) = x + 1 = 0
\n\u21d2 x = – 1
\n\u0905\u092c \u0917\u0941\u0923\u0928\u0916\u0923\u094d\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947
\np(x) = 0 \u092f\u0926\u093f x = -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0905\u0924: p(x) \u092e\u0947\u0902 x = -1 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(x) = x4<\/sup>\u00a0+ x3<\/sup>\u00a0+ x2<\/sup>\u00a0+ x + 1
\np(-1) = (-1)4<\/sup>\u00a0+ (-1)3<\/sup>\u00a0+ (-1)2<\/sup>\u00a0+ (-1) + 1
\n= 1 – 1 + 1 – 1 + 1 = 1
\n\u091a\u0942\u0901\u0915\u093f p(-1) = 1 \u0907\u0938\u0932\u093f\u090f -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0928\u0939\u0940\u0902 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947 x + 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0928\u0939\u0940\u0902 \u0939\u0948 |<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iii) p(x) = x4<\/sup>\u00a0+ 3x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ x + 1
\n\u092e\u093e\u0928\u093e g(x) = x + 1 = 0
\n\u21d2 x = – 1
\n\u0905\u092c \u0917\u0941\u0923\u0928\u0916\u0923\u094d\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947
\np(x) = 0 \u092f\u0926\u093f x = -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0905\u0924: p(x) \u092e\u0947\u0902 x = -1 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(x) = x4<\/sup>\u00a0+ 3x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ x + 1
\np(-1) = (-1)4<\/sup>\u00a0+ 3(-1)3<\/sup>\u00a0+ 3(-1)2<\/sup>\u00a0+ (-1) + 1
\n= 1 – 3 + 3 – 1 + 1 = 1
\n\u091a\u0942\u0901\u0915\u093f p(-1) = 1 \u0907\u0938\u0932\u093f\u090f -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0928\u0939\u0940\u0902 \u0939\u0948 \u0905\u0924: \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947 x + 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0928\u0939\u0940\u0902 \u0939\u0948 |
\n\u092e\u093e\u0928\u093e g(x) = x + 1 = 0
\n\u21d2 x = – 1
\n\u0905\u092c \u0917\u0941\u0923\u0928\u0916\u0923\u094d\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947
\np(x) = 0 \u092f\u0926\u093f x = -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0905\u0924: p(x) \u092e\u0947\u0902 x = -1 \u0930\u0916\u0928\u0947 \u092a\u0930
\n\u0907\u0938\u0932\u093f\u090f -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0928\u0939\u0940\u0902 \u0939\u0948 \u0905\u0924: \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947 x + 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0928\u0939\u0940\u0902 \u0939\u0948 |
\n\"NCERT
\n\"9th<\/p>\n

Ex 2.4 Class 9 \u0917\u0923\u093f\u0924\u00a0Q2. \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0932\u093e\u0917\u0941 \u0915\u0930\u0915\u0947 \u092c\u0924\u093e\u0907\u090f \u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u094d\u0925\u093f\u0924\u093f\u092f\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0938\u094d\u0925\u093f\u0924\u093f \u092e\u0947\u0902 g(x), p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 \u092f\u093e \u0928\u0939\u0940\u0902 :<\/strong>
\n(i) p(x) = 2x3<\/sup>\u00a0+ x2<\/sup>\u00a0– 2x – 1, g(x) = x + 1<\/strong>
\n(ii) p(x) = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1, g(x) = x + 2<\/strong>
\n(iii) p(x) = x3<\/sup>\u00a0– 4x2<\/sup>\u00a0+ x + 6, g(x) = x – 3<\/strong>
\n\u0939\u0932 :<\/strong>\u00a0(i) p(x) = 2x3<\/sup>\u00a0+ x2<\/sup>\u00a0– 2x – 1, g(x) = x + 1
\ng(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915
\n\u21d2\u00a0 x + 1 = 0
\n\u0905\u0924: x = – 1
\n\u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0932\u093e\u0917\u0941 \u0915\u0930\u0928\u0947 \u092a\u0930 \u092f\u0926\u093f p(-1) = 0, \u0924\u094b \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 \u0905\u0925\u0935\u093e \u0928\u0939\u0940\u0902 |
\n\u0905\u0924: p(x) = 2x3<\/sup>\u00a0+ x2<\/sup>\u00a0\u2013 2x \u2013 1\u00a0 \u0926\u093f\u092f\u093e \u0939\u0948 |
\n\u0905\u092c, p(-1) = 2(-1)3<\/sup>\u00a0+ (-1)2<\/sup>\u00a0\u2013 2(-1) \u2013 1
\n= 2 (-1) + 1 + 2 – 1 = – 2 + 1 + 2 – 1 = 0
\n\u091a\u0942\u0901\u0915\u093f p(-1) = 0 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f -1 p(x) \u0915\u093e \u090f\u0915 \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 \u0905\u0924: \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947 x + 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(ii) p(x) = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1, g(x) = x + 2
\ng(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915
\n\u21d2 x + 2 = 0
\n\u0905\u0924: x = – 2
\n\u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0932\u093e\u0917\u0941 \u0915\u0930\u0928\u0947 \u092a\u0930 \u092f\u0926\u093f p(-2) = 0, \u0924\u094b \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 \u0905\u0925\u0935\u093e \u0928\u0939\u0940\u0902 |
\n\u0905\u0924: p(x) = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3x + 1\u00a0 \u0926\u093f\u092f\u093e \u0939\u0948 |
\n\u0905\u092c, p(-2) = (-2)3<\/sup>\u00a0+ 3(-2)2<\/sup>\u00a0+ 3(-2) + 1
\n= -8 + 12 – 6 + 1 = 13 – 14 = – 1
\n\u091a\u0942\u0901\u0915\u093f p(-2) = – 1 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f -2 p(x) \u0915\u093e \u090f\u0915 \u0936\u0941\u0928\u094d\u092f\u0915 \u0928\u0939\u0940\u0902 \u0939\u0948 \u0905\u0924: \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947 x + 2 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092d\u0940 \u0928\u0939\u0940\u0902 \u0939\u0948 |<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iii) p(x) = x3<\/sup>\u00a0\u2013 4x2<\/sup>\u00a0+ x + 6, g(x) = x \u2013 3
\ng(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915
\n\u21d2 x – 3 = 0
\n\u0905\u0924: x = 3
\n\u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0932\u093e\u0917\u0941 \u0915\u0930\u0928\u0947 \u092a\u0930 \u092f\u0926\u093f p(3) = 0, \u0924\u094b \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 \u0905\u0925\u0935\u093e \u0928\u0939\u0940\u0902 |
\n\u0905\u0924: p(x) = x3<\/sup>\u00a0– 4x2<\/sup>\u00a0+ x + 6\u00a0 \u0926\u093f\u092f\u093e \u0939\u0948 |
\n\u0905\u092c, p(3) = (3)3<\/sup>\u00a0– 4(3)2<\/sup>\u00a0+ 3 + 6
\n= 27 – 36 + 3 + 6 = 36 – 36 = 0
\n\u091a\u0942\u0901\u0915\u093f p(3) = 0 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f 3 p(x) \u0915\u093e \u090f\u0915 \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 \u0905\u0924: \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u092e\u0947\u092f \u0938\u0947 x – 3 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

Ex 2.4 Class 9 \u0917\u0923\u093f\u0924\u00a0Q3. k \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f \u091c\u092c\u0915\u093f \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0938\u094d\u0925\u093f\u0924\u093f\u092f\u094b\u0902 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0938\u094d\u0925\u093f\u0924\u093f \u092e\u0947\u0902 (x – 1), p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u094b :<\/strong>
\n(i) p(x) = x2<\/sup>\u00a0+ x + k<\/strong>
\n(ii) p(x) = 2x2<\/sup>\u00a0+ kx +\u00a0\u221a2<\/strong>
\n(iii) p(x) = kx2<\/sup>\u00a0\u2013\u00a0\u221a2x + 1<\/strong>
\n(iv) p(x) = kx2<\/sup>\u00a0\u2013 3x + k<\/strong>
\n\u0939\u0932 :<\/strong>\u00a0(i) p(x) = x2<\/sup>\u00a0+ x + k
\nx – 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f x – 1 = 0 => x = 1
\n\u0905\u0924: 1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f p(1) = 0
\n\u0905\u092c p(x) = x2<\/sup>\u00a0+ x + k = 0
\np(1) = (1)2<\/sup>\u00a0+ (1) + k = 0
\n1 + 1 + k = 0
\n2 + k = 0
\nk = – 2<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(ii) p(x) = 2x2<\/sup>\u00a0+ kx +\u00a0\u221a2
\n\u091a\u0942\u0901\u0915\u093f x – 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948|
\n\u0907\u0938\u0932\u093f\u090f x – 1 = 0
\n\u21d2 x = 1
\n\u0905\u0924: 1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f p(1) = 0
\n\u0905\u092c p(x) = 2x2<\/sup>\u00a0+ kx +\u00a0\u221a2\u00a0= 0
\np(1) = 2(1)2<\/sup>\u00a0+ k(1) +\u00a0\u221a2\u00a0 = 0
\n2 + k +\u00a0\u221a2\u00a0= 0
\nk = – 2 –\u00a0\u221a2
\nk = – (2 +\u00a0\u221a2)<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iii) p(x) = kx2<\/sup>\u00a0\u2013\u00a0\u221a2x + 1
\n\u091a\u0942\u0901\u0915\u093f x – 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f x – 1 = 0 => x = 1
\n\u0905\u0924: 1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f p(1) = 0
\n\u0905\u092c p(x) = kx2<\/sup>\u00a0\u2013\u00a0\u221a2x + 1 = 0
\np(1) = k(1)2<\/sup>\u00a0–\u00a0\u221a2(1) + 1 = 0
\nk –\u00a0\u221a2\u00a0+ 1 = 0
\nk =\u00a0\u221a2\u00a0– 1<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iv) p(x) = kx2<\/sup>\u00a0– 3x + k
\n\u091a\u0942\u0901\u0915\u093f x – 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f x – 1 = 0
\n\u21d2 x = 1
\n\u0905\u0924: 1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 |
\n\u0907\u0938\u0932\u093f\u090f p(1) = 0
\n\u0905\u092c p(x) = kx2<\/sup>\u00a0\u2013 3x + k = 0
\np(1) = k(1)2<\/sup>\u00a0– 3(1) + k = 0
\nk – 3 + k = 0
\n2k – 3 = 0
\n2k = 3
\nk = 3\/2<\/p>\n

Ex 2.4 Class 9 \u0917\u0923\u093f\u0924\u00a0Q4. \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
\n(i) 12x2<\/sup>\u00a0\u2013 7x + 1<\/strong>
\n(ii) 2x2<\/sup>\u00a0+ 7x + 3<\/strong>
\n(iii) 6x2<\/sup>\u00a0+ 5x \u2013 6<\/strong>
\n(iv) 3x2<\/sup>\u00a0\u2013 x \u2013 4<\/strong>
\n\u0939\u0932 :<\/strong>\u00a0(i) 12x2<\/sup>\u00a0\u2013 7x + 1
\n\u21d2 12x2<\/sup>\u00a0– 3x – 4x + 1
\n\u21d2 3x(4x – 1) – 1(4x – 1)
\n\u21d2 (4x – 1) (3x – 1)<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(ii) 2x2<\/sup>\u00a0+ 7x + 3
\n\u21d2 2x2<\/sup>\u00a0+ 6x + x + 3
\n\u21d2 2x(x + 3) + 1(x + 3)
\n\u21d2 (x + 3) (2x + 1)<\/p>\n

\u0939\u0932 :<\/strong>\u00a0\u00a0(iii) 6x2<\/sup>\u00a0+ 5x \u2013 6
\n\u21d2 6x2<\/sup>\u00a0+ 9x – 4x – 6
\n\u21d2 3x(2x + 3) – 2(2x + 3)
\n\u21d2 (2x + 3) (3x – 2)<\/p>\n

\u0939\u0932 :\u00a0<\/strong>(iv) 3x2<\/sup>\u00a0– x – 4
\n\u21d2 3x2<\/sup>\u00a0– 4x + 3x – 4
\n\u21d2 x(3x – 4) + 1(3x – 4)
\n\u21d2 (3x – 4) (x + 1)<\/p>\n

Ex 2.4 Class 9 \u0917\u0923\u093f\u0924\u00a0Q5. \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
\n(i) x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2<\/strong>
\n(ii) x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0\u2013 9x \u2013 5<\/strong>
\n(iii) x3<\/sup>\u00a0+ 13x2<\/sup>\u00a0+ 32x + 20<\/strong>
\n(iv) 2y3<\/sup>\u00a0+ y2<\/sup>\u00a0\u2013 2y \u2013 1<\/strong>
\n\u0939\u0932 :<\/strong>\u00a0(i) x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0938\u0902\u092d\u093e\u0935\u093f\u0924 \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948\u0902 – \u00b11 \u0914\u0930 \u00b12
\n\u0905\u0924: \u092c\u0939\u0941\u092a\u0926 x3<\/sup>\u00a0– 2x2<\/sup>\u00a0– x + 2 \u092e\u0947\u0902 x = 1 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(x) = (1)3<\/sup>\u00a0– 2(1)2<\/sup>\u00a0– (1) + 2
\n=\u00a0 1 – 2 – 1 + 2 = \u00a00
\n\u091a\u0942\u0901\u0915\u093f p(x) = 0 \u0939\u0948, \u0905\u0924: 1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f x – 1 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

\u092a\u0939\u0932\u0940 \u0935\u093f\u0927\u093f :<\/strong>\u00a0x – 1 \u0938\u0947 x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2 \u092e\u0947\u0902 \u092d\u093e\u0917 \u0926\u0947\u0928\u0947 \u092a\u0930
\n\"CBSE
\n\u0905\u0924: x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2 = (x – 1) (x2<\/sup>\u00a0– x – 2) [\u091a\u0942\u0901\u0915\u093f p(x) = g(x) \u00d7 q(x) ]
\n= (x – 1) (x2\u00a0<\/sup>– 2x + x – 2)
\n= (x – 1) [x(x – 2) + 1(x – 2)]
\n= (x – 1) (x – 2) (x + 1)<\/p>\n

\u0928\u094b\u091f:<\/strong>\u00a0\u091a\u0942\u0901\u0915\u093f \u092f\u0939 \u0924\u094d\u0930\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f \u0907\u0938\u0915\u0947 \u0924\u0940\u0928 \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u094b\u0902\u0917\u0947 \u0914\u0930 \u0924\u0940\u0928 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u094b\u0902\u0917\u0947 |<\/p>\n

\u0926\u0942\u0938\u0930\u0940 \u0935\u093f\u0927\u093f :<\/strong>\u00a0\u0939\u092e \u092f\u0939\u093e\u0901 \u092a\u0930 x – 1 \u0938\u0947 \u092d\u093e\u0917 \u0915\u0940 \u0932\u0902\u092c\u0940 \u092a\u094d\u0930\u0915\u094d\u0930\u093f\u092f\u093e \u0928 \u0905\u092a\u0928\u093e\u0915\u0930 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0935\u093f\u0927\u093f \u0938\u0947 \u0905\u0928\u094d\u092f \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0915\u0930 \u0938\u0915\u0924\u0947 \u0939\u0948\u0902 | \u091a\u0942\u0901\u0915\u093f \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 x – 1 \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0939\u0948|
\nx3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2 = x2<\/sup>(x -1) – x2<\/sup>\u00a0– x + 2
\n= x2<\/sup>(x -1) – x(x – 1) – 2x + 2
\n= x2<\/sup>(x -1) – x(x – 1) – 2(x – 1)
\n= (x – 1) (x2<\/sup>\u00a0– x – 2)
\n= (x – 1) (x2\u00a0<\/sup>– 2x + x – 2)
\n= (x – 1) [x(x – 2) + 1(x – 2)]
\n= (x – 1) (x – 2) (x + 1)<\/p>\n

\u0924\u0940\u0938\u0930\u0940 \u0935\u093f\u0927\u093f :<\/strong>\u00a0\u0939\u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0938\u0902\u092d\u093e\u0935\u093f\u0924 \u0936\u0941\u0928\u094d\u092f\u0915 \u00b11 \u0914\u0930 \u00b12 \u091c\u094d\u091e\u093e\u0924 \u0939\u0948 :
\np(x) \u092e\u0947\u0902 x = 1, – 1, 2 \u0914\u0930 – 2 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(1) = 0 \u0939\u0948 | \u0905\u0924: x – 1 \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

\u0905\u092c p(-1) = x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2
\n= (-1)3<\/sup>\u00a0– 2(-1)2<\/sup>\u00a0-(-1) + 2
\n= -1 – 2 + 1 + 2 = 0
\n\u0905\u0924: p(-1) = 0 \u0939\u0948 \u0905\u0924: x + 1 \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

\u0905\u092c p(2) = x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2
\n= (2)3<\/sup>\u00a0– 2(2)2<\/sup>\u00a0-(2) + 2
\n= 8 – 8 – 2 + 2
\n= 0
\np(2) = 0 \u0939\u0948 \u0905\u0924: x – 2 p(x) \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |<\/p>\n

\u0905\u092c p(-2) = x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2
\n= (-2)3<\/sup>\u00a0– 2(-2)2<\/sup>\u00a0-(-2) + 2
\n= -8 – 8 + 2 + 2
\n= -16 + 4 = -12
\np(-2) \u2260 0 \u0905\u0924: – 2 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0928\u0939\u0940\u0902 \u0939\u0948 |
\n\u0905\u0924: \u00a0x3<\/sup>\u00a0\u2013 2x2<\/sup>\u00a0\u2013 x + 2 \u0915\u0947 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 (x – 1) (x + 1) (x – 2)<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(ii) x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0\u2013 9x \u2013 5
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0938\u0902\u092d\u093e\u0935\u093f\u0924 \u0936\u0941\u0928\u094d\u092f\u0915 \u00b1 1 \u0914\u0930 \u00b15 \u0939\u0948 |
\n\u092c\u0939\u0941\u092a\u0926 \u092e\u0947\u0902 x = -1 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(-1) = x3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0\u2013 9x \u2013 5
\n= (-1)3<\/sup>\u00a0\u2013 3(-1)2<\/sup>\u00a0\u2013 9(-1) \u2013 5
\n= -1 \u2013 3 + 9 \u2013 5 = 9 \u2013 9 = 0
\n\u0905\u0924: x = -1 p(x) \u0915\u093e \u0936\u0941\u0928\u094d\u092f\u0915 \u0939\u0948 \u0907\u0938\u0932\u093f\u090f x + 1 \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |
\nx3<\/sup>\u00a0\u2013 3x2<\/sup>\u00a0\u2013 9x \u2013 5 = x2<\/sup>(x + 1) – 4x2<\/sup>\u00a0– 9x – 5
\n= x2<\/sup>(x + 1) – 4x(x + 1) – 5x – 5
\n= x2<\/sup>(x + 1) – 4x(x + 1) – 5(x + 1)
\n= (x + 1) (x2<\/sup>\u00a0– 4x – 5)
\n= (x + 1) (x2<\/sup>\u00a0– 5x + x – 5)
\n= (x + 1) [x(x – 5) +1(x – 5)]
\n= (x + 1) (x – 5) (x + 1)
\n\u0905\u0924: \u0924\u094d\u0930\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926 \u0915\u0947 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 (x + 1), (x – 5) \u0914\u0930 (x + 1) \u0939\u0948 |<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iii) x3<\/sup>\u00a0+ 13x2<\/sup>\u00a0+ 32x + 20
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u0938\u0902\u092d\u093e\u0935\u093f\u0924 \u0936\u0941\u0928\u094d\u092f\u0915 \u00b11, \u00b12, \u00b14, \u00b15, \u00b110 \u0914\u0930 \u00b120 \u0939\u0948\u0902 |
\n\u092c\u0939\u0941\u092a\u0926 \u092e\u0947\u0902 x = – 1 \u0930\u0916\u0928\u0947 \u092a\u0930
\np(x) = x3<\/sup>\u00a0+ 13x2<\/sup>\u00a0+ 32x + 20
\n= (-1)3<\/sup>\u00a0+ 13(-1)2<\/sup>\u00a0+ 32(-1) + 20
\n= -1 + 13 – 32 + 20 = 33 – 33 = 0
\n\u091a\u0942\u0901\u0915\u093f p(-1) = 0 \u0939\u0948 \u0905\u0924: x + 1 \u092c\u0939\u0941\u092a\u0926 \u0915\u093e \u090f\u0915 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u0948 |
\nx3<\/sup>\u00a0+ 13x2<\/sup>\u00a0+ 32x + 20 = x2<\/sup>(x + 1) + 12x2<\/sup>\u00a0+ 32x + 20
\n= x2<\/sup>(x + 1) + 12x(x + 1)\u00a0+ 20x + 20
\n= x2<\/sup>(x + 1) + 12x(x + 1)\u00a0+ 20(x + 1)
\n= (x + 1) (x2<\/sup>\u00a0+ 12x + 20)
\n= (x + 1) (x2<\/sup>\u00a0+ 10x + 2x + 20)
\n= (x + 1) [(x(x + 10) + 2(x + 10)]
\n= (x + 1) (x + 10) (x + 2)
\n\u0905\u0924: \u0924\u094d\u0930\u093f\u0918\u093e\u0924 \u092c\u0939\u0941\u092a\u0926 \u0915\u0947 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 (x + 1), (x + 10) \u0914\u0930 (x + 2) \u0939\u0948|<\/p>\n

\u0939\u0932 :<\/strong>\u00a0(iv) 2y3<\/sup>\u00a0+ y2<\/sup>\u00a0– 2y – 1
\n= y2<\/sup>(2y + 1) -1(2y + 1)
\n= (y2<\/sup>\u00a0– 1) (2y + 1)
\n= (y + 1) ( y – 1) (2y + 1)
\n\u092c\u0939\u0941\u092a\u0926 \u0915\u0947 \u0917\u0941\u0923\u0928\u0916\u0902\u0921 (y + 1), ( y – 1) \u0914\u0930 (2y + 1)\u0939\u0948\u0902 |<\/p>\n

\u0909\u092a\u092f\u094b\u0917\u0940 \u092c\u0940\u091c\u0917\u0923\u093f\u0924\u0940\u092f \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u090f\u0901:\u00a0<\/strong><\/p>\n

    \n
  1. (x + y)2<\/sup>\u00a0= x2<\/sup>\u00a0+ 2xy + y2<\/sup><\/li>\n
  2. (x –\u00a0y)2<\/sup>\u00a0= x2<\/sup>\u00a0–\u00a02xy + y2<\/sup><\/li>\n
  3. x2<\/sup>\u00a0– y2<\/sup>\u00a0= (x + y) (x – y)<\/li>\n
  4. (x + a) (x + b) = x2<\/sup>\u00a0+ (a + b)x + ab<\/li>\n
  5. (x + y)3<\/sup>\u00a0= x3<\/sup>\u00a0+ 3x2<\/sup>y + 3xy2<\/sup>\u00a0+ y3<\/sup><\/li>\n
  6. (x –\u00a0y)3<\/sup>\u00a0= x3<\/sup>\u00a0–\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0–\u00a0y3<\/sup><\/li>\n
  7. x3<\/sup>\u00a0+ y3<\/sup>\u00a0= (x + y) (x2<\/sup>\u00a0–\u00a0xy + y2<\/sup>)<\/li>\n
  8. x3<\/sup>\u00a0–\u00a0y3<\/sup>\u00a0= (x –\u00a0y) (x2<\/sup>\u00a0+\u00a0xy + y2<\/sup>)<\/li>\n
  9. (x + y + z)2<\/sup>\u00a0= x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0+ 2xy + 2yz + 2zx<\/li>\n
  10. x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3\u00a0<\/sup>– 3xyz = ( x + y + z) (x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0–\u00a0xy –\u00a0yz –\u00a0zx)<\/li>\n<\/ol>\n

    \u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.5<\/strong><\/span><\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924 Q1. \u0909\u092a\u092f\u0941\u0915\u094d\u0924 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u0913\u0902 \u0915\u094b \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0915\u0947 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0917\u0941\u0923\u0928\u092b\u0932 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f:<\/strong>
    \n(i) (x + 4) (x + 10)<\/strong>
    \n(ii) (x + 8) (x \u2013 10)<\/strong>
    \n(iii) (3x + 4) (3x \u2013 5)<\/strong>
    \n\"NCERT
    \n(v) (3 \u2013 2x) (3 + 2x)<\/strong>
    \n\u0939\u0932:\u00a0<\/strong>
    \n(i)\u00a0(x + 4) (x + 10)\u00a0<\/strong>
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u00a0(x + a) (x + b) = x2<\/sup>\u00a0+ (a + b)x + ab \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n(x + 4) (x + 10)\u00a0=\u00a0x2<\/sup>\u00a0+ (4\u00a0+ 10)x + (4)(10)
    \n=\u00a0x2<\/sup>\u00a0+ 14x + 40<\/p>\n

    (ii) (x + 8) (x \u2013 10)\u00a0<\/strong>
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u00a0(x + a) (x + b) = x2<\/sup>\u00a0+ (a + b)x + ab \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n(x + 8) (x \u2013 10)\u00a0=\u00a0x2<\/sup>\u00a0+ [8\u00a0+ (-10)]x + (8)(-10)
    \n=\u00a0x2<\/sup>\u00a0– 2x – 80<\/p>\n

    (iii) (3x + 4) (3x \u2013 5)<\/strong>
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u00a0(x + a) (x + b) = x2<\/sup>\u00a0+ (a + b)x + ab \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n(3x + 4) (3x \u2013 5) = (3x)2<\/sup>\u00a0+ [4\u00a0+ (-5)]3x + (4)(-5)
    \n= 9x2<\/sup>\u00a0–\u00a03x – 20
    \n\"NCERT
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u00a0(x + y) (x – y)\u00a0=\u00a0x2<\/sup>\u00a0– y2<\/sup>\u00a0\u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n\"NCERT<\/p>\n

    (v) (3 \u2013 2x) (3 + 2x)<\/strong>
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u00a0(x + y) (x – y)\u00a0=\u00a0x2<\/sup>\u00a0– y2<\/sup>\u00a0\u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930\u00a0\u00a0\u200b
    \n(3 \u2013 2x) (3 + 2x) = (3)2<\/sup>\u00a0– (2x)2<\/sup>
    \n= 9 – 4x2<\/sup><\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q2. \u0938\u0940\u0927\u0947 \u0917\u0941\u0928\u093e \u0915\u093f\u092f\u0947 \u092c\u093f\u0928\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0917\u0941\u0923\u0928\u092b\u0932\u094b\u0902 \u0915\u0947 \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
    \n(i) 103 \u00d7 107<\/strong>
    \n(ii) 95 \u00d7 96<\/strong>
    \n(iii) 104 \u00d7 96<\/strong>
    \n\u0939\u0932:<\/strong>
    \n(i) 103 \u00d7 107\u00a0= (100 + 3) (100 + 7)
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u00a0(x + a) (x + b) = x2<\/sup>\u00a0+ (a + b)x + ab \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n(100 + 3) (100 + 7)\u00a0= (100)2<\/sup>\u200b + (3 + 7)100 + 3\u00d77
    \n=10000 + 1000 + 21 = 11021<\/p>\n

    (ii) 95\u00a0\u00d7 96\u00a0= (90\u00a0+ 5) (90\u00a0+ 6)
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u00a0(x + a) (x + b) = x2<\/sup>\u00a0+ (a + b)x + ab \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n(90\u00a0+ 5) (90\u00a0+ 6)\u00a0= (90)2<\/sup>\u200b + (5\u00a0+ 6)90 + 5\u00d76
    \n= 8100\u00a0+ 990\u00a0+ 30 = 9120<\/p>\n

    (iii) \u00a0104\u00a0\u00d7 96\u00a0= (100\u00a0+ 4) (100\u00a0– 4)
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u00a0(x + y) (x – y)\u00a0=\u00a0x2<\/sup>\u00a0– y2<\/sup>\u00a0\u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0928\u0947 \u092a\u0930\u00a0\u00a0\u200b
    \n(100)2<\/sup>\u00a0– (4)2<\/sup>
    \n= 10000 – 16 = 9984<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a03. \u0909\u092a\u092f\u0941\u0915\u094d\u0924 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u0913\u0902 \u0915\u094b \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0915\u0947 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u093e \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0915\u0940\u091c\u093f\u090f:<\/strong>
    \n(i) 9x2<\/sup>\u00a0+ 6xy + y2<\/sup>\u00a0<\/strong>
    \n(ii) 4y2<\/sup>\u00a0\u2013 4y + 1<\/strong>
    \n\"NCERT
    \n\u0939\u0932:<\/strong>
    \n(i) 9x2<\/sup>\u00a0+ 6xy + y2<\/sup>\u00a0<\/strong>
    \n= (3x)2<\/sup>\u00a0+ 2.3x.y + (y)2 \u00a0 \u00a0<\/sup>\u00a0[\u00a0\u2235 x2<\/sup>\u00a0+ 2xy + y2<\/sup>\u00a0= (x + y)2<\/sup>]
    \n\u2234 = (3x + y)2<\/sup>
    \n= \u00a0(3x + y)\u00a0\u00a0(3x + y)<\/p>\n

    (ii) 4y2<\/sup>\u00a0– 4y + 1\u00a0<\/strong>
    \n= (2y)2<\/sup>\u00a0–\u00a02.2y.1\u00a0+ (1)2 \u00a0 \u00a0<\/sup>\u00a0[\u00a0\u2235 x2<\/sup>\u00a0–\u00a02xy + y2<\/sup>\u00a0= (x –\u00a0y)2<\/sup>]
    \n\u2234 = (2y\u00a0–\u00a01)2<\/sup>
    \n= \u00a0(2y – 1)\u00a0\u00a0(2y – 1)
    \n\"NCERT
    \n\"NCERT
    \n[\u00a0\u2235\u00a0x2<\/sup>\u00a0– y2<\/sup>\u00a0= (x + y) (x – y)\u00a0\u200b]<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q4. \u0909\u092a\u092f\u0941\u0915\u094d\u0924 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e\u0913\u0902 \u0915\u094b \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930\u0915\u0947 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u092a\u094d\u0930\u0938\u093e\u0930 \u0915\u0940\u091c\u093f\u090f:<\/strong>
    \n(i) (x + 2y + 4z)2<\/sup>\u00a0<\/strong>
    \n(ii) (2x \u2013 y + z)2<\/sup>\u00a0<\/strong>
    \n(iii) (\u20132x + 3y + 2z)2<\/sup><\/strong>
    \n(iv) (3a \u2013 7b \u2013 c)2<\/sup>\u00a0<\/strong>
    \n(v) (\u20132x + 5y \u2013 3z)2<\/sup><\/strong>
    \n\u0939\u0932:<\/strong>
    \n(i) (x + 2y + 4z)2<\/sup>\u00a0\u00a0<\/strong>
    \n\u092f\u0939\u093e\u0901 \u092e\u093e\u0928\u093e \u0915\u093f a = x, b = 2y, c = 4z \u0914\u0930 a, b \u0924\u0925\u093e\u00a0c \u0915\u093e \u092e\u093e\u0928 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e
    \n(a\u00a0+ b\u00a0+ c)2<\/sup>\u00a0= a2<\/sup>\u00a0+ b2<\/sup>\u00a0+ c2<\/sup>\u00a0+ 2ab\u00a0+ 2bc\u00a0+ 2ca \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930
    \n\u2234\u00a0(x + 2y + 4z)2<\/sup>\u00a0= (x)2<\/sup>\u00a0+ (2y)2<\/sup>\u00a0+ (4z)2<\/sup>\u00a0+ 2(x)(2y)\u00a0+ 2(2y)(4z)\u00a0+ 2(4z)(x)
    \n= x2<\/sup>\u00a0+ 4y2<\/sup>\u00a0+ 16z2<\/sup>\u00a0+ 4xy + 16yz + 8zx<\/p>\n

    (ii) (2x \u2013 y + z)2<\/sup>\u00a0<\/strong>
    \n\u092f\u0939\u093e\u0901 \u092e\u093e\u0928\u093e \u0915\u093f\u00a0a = 2x, b = –\u00a0y, c = z\u00a0\u0914\u0930 a, b \u0924\u0925\u093e\u00a0c \u0915\u093e \u092e\u093e\u0928 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e
    \n(a\u00a0+ b\u00a0+ c)2<\/sup>\u00a0= a2<\/sup>\u00a0+ b2<\/sup>\u00a0+ c2<\/sup>\u00a0+ 2ab\u00a0+ 2bc\u00a0+ 2ca \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930
    \n\u2234\u00a0(2x \u2013 y + z)2<\/sup>\u00a0= (2x)2<\/sup>\u00a0+ (-\u00a0y)2<\/sup>\u00a0+ (z)2<\/sup>\u00a0+ 2(2x)(-\u00a0y)\u00a0+ 2(-\u00a0y)(z)\u00a0+ 2(z)(2x)
    \n= 4x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0–\u00a04xy –\u00a02yz + 4zx<\/p>\n

    (iii) (\u20132x + 3y + 2z)2<\/sup><\/strong>
    \n\u092f\u0939\u093e\u0901 \u092e\u093e\u0928\u093e \u0915\u093f\u00a0a = – 2x, b = 3y, c = 2z\u00a0\u0914\u0930 a, b \u0924\u0925\u093e\u00a0c \u0915\u093e \u092e\u093e\u0928 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e
    \n(a\u00a0+ b\u00a0+ c)2<\/sup>\u00a0= a2<\/sup>\u00a0+ b2<\/sup>\u00a0+ c2<\/sup>\u00a0+ 2ab\u00a0+ 2bc\u00a0+ 2ca \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930
    \n\u2234\u00a0(-2x + 3y + 2z)2<\/sup>
    \n= (-2x)2<\/sup>\u00a0+ (3y)2<\/sup>\u00a0+ (2z)2<\/sup>\u00a0+ 2(-2x)(3y)\u00a0+ 2(3y)(2z)\u00a0+ 2(2z)(-2x)
    \n= 4x2<\/sup>\u00a0+ 9y2<\/sup>\u00a0+ 4z2<\/sup>\u00a0–\u00a012xy \u00a0+ 12yz\u00a0– 8zx<\/p>\n

    (iv) (3a – 7b – c)2<\/sup>\u00a0<\/strong>
    \n\u092f\u0939\u093e\u0901 \u092e\u093e\u0928\u093e \u0915\u093f\u00a0x\u00a0= 3a, y\u00a0=\u00a0-7b, z\u00a0=\u00a0-c \u0914\u0930 x, y\u00a0\u0924\u0925\u093e z\u00a0\u0915\u093e \u092e\u093e\u0928 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e
    \n(x + y + z)2<\/sup>\u00a0= x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0+ 2xy + 2yz + 2zx\u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930
    \n\u2234\u00a0(3a – 7b – c)2<\/sup>
    \n= (3a)2<\/sup>\u00a0+ (-7b)2<\/sup>\u00a0+ (-c)2<\/sup>\u00a0+ 2(3a)(-7b)\u00a0+ 2(-7b)(-c)\u00a0+ 2(-c)(3a)
    \n= 9a2<\/sup>\u00a0+ 49b2<\/sup>\u00a0+ c2\u00a0<\/sup>–\u00a042ab\u00a0 + 14bc\u00a0– 6ac<\/p>\n

    (v) (-2x + 5y – 3z)2<\/sup><\/strong>
    \n\u092f\u0939\u093e\u0901 \u092e\u093e\u0928\u093e \u0915\u093f\u00a0a = – 2x, b = 5y, c =\u00a0-3z\u00a0\u0914\u0930 a, b \u0924\u0925\u093e\u00a0c \u0915\u093e \u092e\u093e\u0928 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e
    \n(a\u00a0+ b\u00a0+ c)2<\/sup>\u00a0= a2<\/sup>\u00a0+ b2<\/sup>\u00a0+ c2<\/sup>\u00a0+ 2ab\u00a0+ 2bc\u00a0+ 2ca \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930
    \n\u2234\u00a0(-2x + 5y – 3z)2<\/sup>
    \n= (-2x)2<\/sup>\u00a0+ (5y)2<\/sup>\u00a0+ (-3z)2<\/sup>\u00a0+ 2(-2x)(5y)\u00a0+ 2(5y)(-3z)\u00a0+ 2(-3z)(-2x)
    \n= 4x2<\/sup>\u00a0+ 25y2<\/sup>\u00a0+ 9z2<\/sup>\u00a0– 20xy \u00a0–\u00a030yz +\u00a012zx<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q5. \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0915\u0940\u091c\u093f\u090f:
    \n(i) 4x2<\/sup>\u00a0+ 9y2<\/sup>\u00a0+ 16z2<\/sup>\u00a0+ 12xy – 24yz – 16xz<\/strong>
    \n\u0939\u0932:<\/strong>
    \n(i) 4x2<\/sup>\u00a0+ 9y2<\/sup>\u00a0+ 16z2<\/sup>\u00a0+ 12xy – 24yz – 16xz
    \n= (2x)2<\/sup>\u00a0+ (3y)2<\/sup>\u00a0+ (4z)2<\/sup>\u00a0+ 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)
    \n[\u2235\u00a0a2<\/sup>\u00a0+ b2<\/sup>\u00a0+ c2<\/sup>\u00a0+ 2ab\u00a0+ 2bc\u00a0+ 2ca =\u00a0(a\u00a0+ b\u00a0+ c)2<\/sup>\u00a0]
    \n= (2x + 3y + 4z)2<\/sup>
    \n=\u00a0(2x + 3y + 4z)\u00a0(2x + 3y + 4z)
    \n\"NCERT
    \n\"Maths<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q6. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0918\u0928\u094b\u0902 \u0915\u094b \u0935\u093f\u0938\u094d\u0924\u093e\u0930\u093f\u0924 \u0930\u0942\u092a \u092e\u0947\u0902 \u0932\u093f\u0916\u093f\u090f\u00a0:
    \n(i) (2x + 1)3<\/sup>\u00a0<\/strong>
    \n(ii) (2a \u2013 3b)3<\/sup><\/strong>
    \n\"Class
    \n\"NCERT
    \n\u0939\u0932:<\/strong>
    \n(i) (2x + 1)3<\/sup>\u00a0<\/strong>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947 (a\u00a0+ b)3<\/sup>\u00a0= a3<\/sup>\u00a0+ 3a2<\/sup>b\u00a0+ 3ab2<\/sup>\u00a0+ b3<\/sup>]
    \n(2x + 1)3<\/sup>\u00a0= (2x)3<\/sup>\u00a0+ 3 (2x)2\u00a0<\/sup>(1) + 3 (2x) (1)2<\/sup>\u00a0+ (1)3<\/sup>
    \n= 8x3<\/sup>\u00a0+ 12x2<\/sup>\u00a0+ 6x + 1<\/p>\n

    (ii) (2a \u2013 3b)3<\/sup><\/strong>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0(x –\u00a0y)3<\/sup>\u00a0= x3<\/sup>\u00a0–\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0–\u00a0y3<\/sup>]
    \n(2a\u00a0– 3b)3<\/sup>\u00a0= (2a)3<\/sup>\u00a0–\u00a03 (2a)2\u00a0<\/sup>(3b) + 3(2a) (3b)2<\/sup>\u00a0–\u00a0(3b)3<\/sup>
    \n= 8a3<\/sup>\u00a0– 36a2<\/sup>b\u00a0+ 54ab2<\/sup>\u00a0–\u00a027b3<\/sup>
    \n\"NCERT
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0(a\u00a0+ b)3<\/sup>\u00a0= a3<\/sup>\u00a0+ 3a2<\/sup>b\u00a0+ 3ab2<\/sup>\u00a0+ b3<\/sup>]
    \n\"NCERT
    \n\"NCERT
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0(a –\u00a0b)3<\/sup>\u00a0= a3<\/sup>\u00a0–\u00a03a2<\/sup>b\u00a0+ 3ab2<\/sup>\u00a0–\u00a0b3<\/sup>]
    \n\"NCERT<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q7. \u0909\u092a\u092f\u0941\u0915\u094d\u0924 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u093e \u092a\u094d\u0930\u092f\u094b\u0917 \u0915\u0930 \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
    \n(i) (99)3<\/sup>\u00a0<\/strong>
    \n(ii) (102)3<\/sup>\u00a0<\/strong>
    \n(iii) (998)3<\/sup><\/strong>
    \n\u0939\u0932 :\u00a0<\/strong>
    \n(i) (99)3<\/sup>\u00a0<\/strong>
    \n=\u00a0(100 – 1)3<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0(a –\u00a0b)3<\/sup>\u00a0= a3<\/sup>\u00a0–\u00a03a2<\/sup>b\u00a0+ 3ab2<\/sup>\u00a0–\u00a0b3<\/sup>]
    \n(100 – 1)3\u00a0<\/sup>=\u00a0(100)3<\/sup>\u00a0–\u00a03(100)2<\/sup>(1) + 3(100)(1)2<\/sup>\u00a0–\u00a0(1)3<\/sup>
    \n= 1000000\u00a0– 30000\u00a0+ 300\u00a0– 1 = 1000300 – 30001 = 970299<\/p>\n

    (ii) (102)3<\/sup>\u00a0<\/strong>
    \n=\u00a0(100 +\u00a02)3<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0(a\u00a0+ b)3<\/sup>\u00a0= a3<\/sup>\u00a0+ 3a2<\/sup>b\u00a0+ 3ab2<\/sup>\u00a0+ b3<\/sup>]
    \n(100 + 2)3\u00a0<\/sup>=\u00a0(100)3<\/sup>\u00a0+\u00a03 (100)2\u00a0<\/sup>(2)+ 3 (100) (2)2<\/sup>\u00a0+\u00a0(2)3<\/sup>
    \n= 1000000 +\u00a060000\u00a0+ 1200 + 8 = 1061208<\/p>\n

    (iii) (998)3<\/sup><\/strong>
    \n=\u00a0(1000 –\u00a02)3<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0(a –\u00a0b)3<\/sup>\u00a0= a3<\/sup>\u00a0–\u00a03a2<\/sup>b\u00a0+ 3ab2<\/sup>\u00a0–\u00a0b3<\/sup>]
    \n(1000 –\u00a02)3\u00a0<\/sup>=\u00a0(1000)3<\/sup>\u00a0–\u00a03 (1000)2\u00a0<\/sup>(2)+ 3(1000) (2)2<\/sup>\u00a0–\u00a0(2)3<\/sup>
    \n= 1000000000 –\u00a06000000 + 12000 – 8
    \n= 1000012000 – 6000008
    \n= 994011992<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q8. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0915\u093e \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0915\u0940\u091c\u093f\u090f\u00a0:<\/strong>
    \n(i) 8a3<\/sup>\u00a0+ b3\u00a0+ 12a2<\/sup>b + 6ab2<\/sup><\/strong>
    \n(ii) 8a2<\/sup>\u00a0– b2<\/sup>\u00a0– 12a2<\/sup>b + 6ab2<\/sup><\/strong>
    \n(iii) 27 – 125a3<\/sup>\u00a0– 135a + 225a2<\/sup><\/strong>
    \n(iv) 64a3<\/sup>\u00a0– 27b3<\/sup>\u00a0– 144a2<\/sup>b + 108ab2<\/sup><\/strong>
    \n\"NCERT
    \n\u0939\u0932:<\/strong>
    \n(i) 8a3<\/sup>\u00a0+ b3\u00a0+ 12a2<\/sup>b + 6ab2<\/sup>\u00a0<\/strong>
    \n= (2a)3<\/sup>\u00a0+(b)3<\/sup>\u00a0+ 3(2a)2<\/sup>(b) + 3(2a)(b)2<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0\u00a0x3<\/sup>\u00a0\u00a0+\u00a0y3\u00a0<\/sup>+\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0=\u00a0(x +\u00a0y)3<\/sup>\u00a0]
    \n=\u00a0(2a)3<\/sup>\u00a0+(b)3<\/sup>\u00a0+ 3(2a)2<\/sup>(b) + 3(2a)(b)2<\/sup>\u00a0= (2a + b)3<\/sup>
    \n= (2a + b)(2a + b)(2a + b)<\/p>\n

    (ii) 8a2<\/sup>\u00a0– b2<\/sup>\u00a0– 12a2<\/sup>b + 6ab2<\/sup><\/strong>
    \n= (2a)3<\/sup>\u00a0–\u00a0(b)3<\/sup>\u00a0–\u00a03(2a)2<\/sup>(b) + 3(2a)(b)2<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0\u00a0x3<\/sup>\u00a0– y3\u00a0<\/sup>–\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0=\u00a0(x –\u00a0y)3<\/sup>\u00a0]
    \n=\u00a0(2a)3<\/sup>\u00a0–\u00a0(b)3<\/sup>\u00a0–\u00a03(2a)2<\/sup>(b) + 3(2a)(b)2<\/sup>\u00a0= (2a –\u00a0b)3<\/sup>\u200b
    \n= (2a –\u00a0b)(2a –\u00a0b)(2a –\u00a0b)<\/p>\n

    (iii) 27 – 125a3<\/sup>\u00a0– 135a + 225a2<\/sup><\/strong>
    \n= (3)3<\/sup>\u00a0–\u00a0(5a)3<\/sup>\u00a0–\u00a03(3)2<\/sup>(5a) + 3(3)(5a)2<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0\u00a0x3<\/sup>\u00a0– y3\u00a0<\/sup>–\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0=\u00a0(x –\u00a0y)3<\/sup>\u00a0]
    \n= (3)3<\/sup>\u00a0–\u00a0(5a)3<\/sup>\u00a0–\u00a03(3)2<\/sup>(5a) + 3(3)(5a)2<\/sup>= (3\u00a0– 5a)3<\/sup>\u200b
    \n=\u00a0(3\u00a0– 5a)(3\u00a0– 5a)(3\u00a0– 5a)<\/p>\n

    (iv) 64a3\u00a0<\/sup>– 27b3<\/sup>\u00a0– 144a2<\/sup>b + 108ab2<\/sup>\u00a0\u00a0<\/strong>
    \n= (4a)3<\/sup>\u00a0–\u00a0(3b)3<\/sup>\u00a0–\u00a03(4a)2<\/sup>(3b) + 3(4a)(3b)2<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0x3<\/sup>\u00a0– y3\u00a0<\/sup>–\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0=\u00a0(x –\u00a0y)3<\/sup>\u00a0]
    \n=\u00a0(4a)3<\/sup>\u00a0–\u00a0(3b)3<\/sup>\u00a0–\u00a03(4a)2<\/sup>(3b) + 3(4a)(3b)2<\/sup>\u00a0= (4a – 3b)3<\/sup>\u200b
    \n=\u00a0(4a – 3b)(4a – 3b)(4a – 3b)
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0x3<\/sup>\u00a0– y3\u00a0<\/sup>–\u00a03x2<\/sup>y + 3xy2<\/sup>\u00a0=\u00a0(x –\u00a0y)3<\/sup>\u00a0]
    \n\"Maths<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q9. \u0938\u0924\u094d\u092f\u093e\u092a\u093f\u0924 \u0915\u0940\u091c\u093f\u090f :<\/strong>
    \n(i) x3<\/sup>\u00a0+ y3<\/sup>\u00a0= (x + y) (x2<\/sup>\u00a0– xy + y2<\/sup>)<\/strong>
    \n\u0939\u0932 :<\/strong>
    \nRHS =\u00a0(x + y) (x2<\/sup>\u00a0– xy + y2<\/sup>)
    \n= x(x2<\/sup>\u00a0– xy + y2<\/sup>) + y (x2<\/sup>\u00a0– xy + y2<\/sup>)
    \n= x3<\/sup>\u00a0– x2<\/sup>y + xy2<\/sup>\u00a0+ x2<\/sup>y – xy2<\/sup>\u00a0+ y3<\/sup>
    \n\"Maths
    \n= x3<\/sup>\u00a0+ y3<\/sup>
    \n\u2235\u00a0LHS = RHS \u0938\u0924\u094d\u092f\u093e\u092a\u093f\u0924<\/p>\n

    (ii) x3<\/sup>\u00a0– y3<\/sup>\u00a0= (x – y) (x2<\/sup>\u00a0+ xy + y2<\/sup>)<\/strong>
    \n\u0939\u0932 :<\/strong>
    \nRHS = (x – y) (x2<\/sup>\u00a0+ xy + y2<\/sup>)
    \nx(x2<\/sup>\u00a0+ xy + y2<\/sup>) – y (x2<\/sup>\u00a0+ xy + y2<\/sup>)
    \n= x3<\/sup>\u00a0+\u00a0x2<\/sup>y\u00a0+ xy2<\/sup>\u00a0– x2<\/sup>y\u00a0–\u00a0xy2<\/sup>\u00a0– y3<\/sup>
    \n\"Maths
    \n= x3<\/sup>\u00a0– y3<\/sup>
    \n\u2235\u00a0LHS = RHS \u0938\u0924\u094d\u092f\u093e\u092a\u093f\u0924 |<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q10. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f:<\/strong>
    \n(i) 27y3<\/sup>\u00a0+ 125z3<\/sup>\u00a0<\/strong>
    \n(ii) 64m3<\/sup>\u00a0– 343n3<\/sup><\/strong>
    \n\u0939\u0932 :\u00a0<\/strong>
    \n(i) 27y3<\/sup>\u00a0+ 125z3<\/sup>\u00a0<\/strong>
    \n= (3y)3<\/sup>\u00a0+ (5z)3<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0x3<\/sup>\u00a0+ y3<\/sup>\u00a0= (x + y) (x2<\/sup>\u00a0– xy + y2<\/sup>) ]
    \n(3y)3<\/sup>\u00a0+ (5z)3<\/sup>\u200b\u00a0= (3y + 5y) [(3y)2<\/sup>\u00a0– (3y)(5z) + (5z)2<\/sup>]
    \n=\u00a0(3y + 5y)\u00a0(9y2<\/sup>\u00a0– 15yz + 25z2<\/sup>)<\/p>\n

    (ii) 64m3<\/sup>\u00a0– 343n3<\/sup>
    \n\u0939\u0932 :\u00a0<\/strong>
    \n(ii) 64m3<\/sup>\u00a0\u2013 343n3<\/sup><\/strong>
    \n= (4m)3<\/sup>\u00a0–\u00a0(7n)3<\/sup>
    \n[\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0x3<\/sup>\u00a0–\u00a0y3<\/sup>\u00a0= (x\u00a0–\u00a0y) (x2<\/sup>\u00a0+ xy + y2<\/sup>) ]
    \n(4m)3<\/sup>\u00a0–\u00a0(7n)3<\/sup>\u200b\u00a0= (4m\u00a0\u2013\u00a07n) [(4m)2<\/sup>\u00a0+ (4m)(7n) + (7n)2<\/sup>]
    \n=\u00a0(4m\u00a0–\u00a07n)\u00a0(16m2<\/sup>\u00a0+\u00a028mn\u00a0+ 49n\u200b2<\/sup>)<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q11. \u0917\u0941\u0923\u0928\u0916\u0923\u094d\u0921 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f : 27x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0– 9xyz<\/strong>
    \n\u0939\u0932 :\u00a0<\/strong>
    \n= (3x)3<\/sup>\u00a0+ (y)3<\/sup>\u00a0+ (z)3\u00a0<\/sup>– 9xyz
    \n\u2235\u00a0x3\u00a0<\/sup>+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0– 3xyz =\u00a0\u00a0(x + y + z) (x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z\u200b2<\/sup>\u00a0– xy – yz – zx)
    \n\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0:
    \n= (3x + y + z) ((3x)2<\/sup>\u00a0+ (y)2<\/sup>\u00a0+ (z)2<\/sup>\u00a0– (3x)(y) – (y)(z) – (z)(3x))
    \n=\u00a0(3x + y + z) (9x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0– 3xy – yz – 3zx)<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q12. \u0938\u0924\u094d\u092f\u093e\u092a\u093f\u0924 \u0915\u0940\u091c\u093f\u090f:<\/strong>\u00a0<\/strong>
    \nx3\u00a0<\/sup>+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0– 3xyz\u00a0= \\(\\frac { 1 }{ 2 }\\) (x + y + z) [(x – y)2<\/sup>\u00a0+ (y – z)2<\/sup>\u00a0+ (z – x)2<\/sup>]
    \n\u0939\u0932 :\u00a0<\/strong>
    \nLHS =\u00a0 \\(\\frac { 1 }{ 2 }\\) (x + y + z) [x2<\/sup>\u00a0– 2xy + y2<\/sup>\u00a0+ y2<\/sup>\u00a0– 2yz +\u00a0z2<\/sup>\u00a0+ z2<\/sup>\u00a0– 2xz +\u00a0x2<\/sup>]
    \n= \\(\\frac { 1 }{ 2 }\\) (x + y + z) (2x2\u00a0<\/sup>+ 2y2<\/sup>\u00a0+ 2z2<\/sup>\u00a0– 2xy – 2yz – 2xz)
    \n= \\(\\frac { 1 }{ 2 }\\) \u00d7\u00a02(x + y + z) (x2\u00a0<\/sup>+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0– xy – yz – xz)
    \n=\u00a0(x + y + z)(x2\u00a0<\/sup>+\u00a0y2<\/sup>\u00a0+\u00a0z2<\/sup>\u00a0–\u00a0xy –\u00a0yz –\u00a0xz)
    \n=\u00a0x3\u00a0<\/sup>+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0– 3xyz [\u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u0915\u0947 \u092a\u094d\u0930\u092f\u094b\u0917 \u0938\u0947\u00a0]
    \nLHS = RHS<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q13. \u092f\u0926\u093f x + y + z = 0 \u0939\u094b, \u0924\u094b \u0926\u093f\u0916\u093e\u0907\u090f \u0915\u093f x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>= 3xyz \u0939\u0948 | \u200b<\/strong>
    \n\u0939\u0932 :\u00a0<\/strong>x + y + z = 0 \u0926\u093f\u092f\u093e \u0939\u0948 |
    \nx3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>–\u00a03xyz =\u00a0( x + y + z) (x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0–\u00a0xy –\u00a0yz –\u00a0zx)
    \n= (0)\u00a0(x2<\/sup>\u00a0+ y2<\/sup>\u00a0+ z2<\/sup>\u00a0–\u00a0xy –\u00a0yz –\u00a0zx) = 0
    \n\u0905\u0924: x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>–\u00a03xyz = 0
    \n\u092f\u093e x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>=\u00a03xyz\u00a0\u0938\u0924\u094d\u092f\u093e\u092a\u093f\u0924<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q14. \u0935\u093e\u0938\u094d\u0924\u0935 \u092e\u0947\u0902 \u0918\u0928\u094b\u0902 \u0915\u093e \u092a\u0930\u093f\u0915\u0932\u0928 \u0915\u093f\u090f \u092c\u093f\u0928\u093e \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u093e \u092e\u093e\u0928 \u091c\u094d\u091e\u093e\u0924 \u0915\u0940\u091c\u093f\u090f :\u00a0<\/strong>
    \n(i)\u00a0(-12)3<\/sup>\u00a0+ (7)3<\/sup>\u00a0+ (5)3<\/sup><\/strong>
    \n(ii) (28)3<\/sup>\u00a0+ (-15)3<\/sup>\u00a0+ (-13)3<\/sup><\/strong>
    \n\u0939\u0932 :<\/strong>\u00a0(i)\u00a0(-12)3<\/sup>\u00a0+ (7)3<\/sup>\u00a0+ (5)3<\/sup><\/p>\n

    \u092a\u094d\u0930\u0936\u094d\u0928 13. \u092e\u0947\u0902 \u0939\u092e\u0928\u0947 \u090f\u0915 \u0938\u0930\u094d\u0935\u0938\u092e\u0940\u0915\u093e \u092a\u094d\u0930\u093e\u092a\u094d\u0924 \u0915\u093f\u092f\u093e \u0925\u093e \u0915\u093f \u092f\u0926\u093f x + y + z = 0 \u0939\u094b \u0924\u094b
    \nx3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>=\u00a03xyz\u00a0\u0939\u0948 |
    \n\u0905\u0924: \u0907\u0938 \u0938\u0930\u094d\u0935\u0938\u092e\u093f\u0915\u093e \u092e\u0947\u0902 x = -12, y = 7 \u0914\u0930 z = 5 \u0930\u0916\u0928\u0947 \u092a\u0930
    \n\u091a\u0942\u0901\u0915\u093f – 12 + 7 + 5 => -12 + 12 = 0
    \n\u0905\u0924: x + y + z = 0 \u0939\u0948 |
    \n\u0905\u092c, x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>=\u00a03xyz\u00a0 [x, y, \u0914\u0930 z \u0915\u093e \u092e\u093e\u0928 \u0930\u0916\u0928\u0947 \u092a\u0930 ]
    \n=> (-12)3<\/sup>\u00a0+ (7)3<\/sup>\u00a0+ (5)3<\/sup>\u00a0= 3\u00a0\u00d7 (-12)\u00a0\u00d7 7\u00a0\u00d7 5
    \n= – 1260<\/p>\n

    \u0939\u0932 :<\/strong>\u00a0(ii) (28)3<\/sup>\u00a0+ (\u201315)3<\/sup>\u00a0+ (\u201313)3<\/sup>
    \n28 + (-15) + (-13) = 28 – 28 = 0
    \n\u091a\u0942\u0901\u0915\u093f x + y + z = 0 \u0939\u0948 |
    \n\u0907\u0938\u0932\u093f\u090f x3<\/sup>\u00a0+ y3<\/sup>\u00a0+ z3<\/sup>\u00a0<\/sup>=\u00a03xyz
    \n\u0905\u092c,\u00a0(28)3<\/sup>\u00a0+ (\u201315)3<\/sup>\u00a0+ (\u201313)3<\/sup>\u200b = 3\u00a0\u00d7 28\u00a0\u00d7 (-15)\u00a0\u00d7 (-13)
    \n=\u00a0133380<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q15. \u0928\u0940\u091a\u0947 \u0926\u093f\u090f \u0917\u090f \u0906\u092f\u093e\u0924\u094b\u0902, \u091c\u093f\u0928\u092e\u0947\u0902 \u0909\u0928\u0915\u0947 \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 \u0926\u093f\u090f \u0917\u090f \u0939\u0948, \u092e\u0947\u0902 \u0938\u0947 \u092a\u094d\u0930\u0924\u094d\u092f\u0947\u0915 \u0915\u0940 \u0932\u0902\u092c\u093e\u0908 \u0914\u0930 \u091a\u094c\u095c\u093e\u0908 \u0915\u0947 \u0932\u093f\u090f \u0938\u0902\u092d\u0935 \u0935\u094d\u092f\u0902\u091c\u0915 \u0926\u0940\u091c\u093f\u092f\u0947 |\u00a0<\/strong>
    \n(i) \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 : 25a2<\/sup>\u00a0– 35a + 12\u00a0<\/strong>
    \n(ii) \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 : 35y2<\/sup>\u00a0+ 13y – 12\u00a0<\/strong>
    \n\u0939\u0932 :<\/strong>\u00a0(i) \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 : 25a2<\/sup>\u00a0– 35a + 12
    \n\u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 = \u0932\u0902\u092c\u093e\u0908\u00a0\u00d7 \u091a\u094c\u095c\u093e\u0908
    \n\u0905\u0924:\u00a025a2<\/sup>\u00a0– 35a + 12 \u0915\u0947 \u0926\u094b \u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0939\u094b\u0902\u0917\u0947 \u091c\u093f\u0938\u092e\u0947\u0902 \u090f\u0915 \u0932\u0902\u092c\u093e\u0908 \u0939\u094b\u0917\u093e \u0914\u0930 \u0926\u0942\u0938\u0930\u093e \u091a\u094c\u095c\u093e\u0908 \u0939\u094b\u0917\u093e |
    \n\u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0915\u0930\u0928\u0947 \u092a\u0930 :
    \n25a2<\/sup>\u00a0– 35a + 12\u00a0= 25a2<\/sup>\u00a0+ 15a + 20a + 12
    \n= 5a(5a + 3) + 4(5a + 3)
    \n= (5a + 3) (5a + 4)
    \n\u091a\u0942\u0901\u0915\u093f (5a + 3) < (5a + 4) \u0939\u0948 |
    \n\u0905\u0924: \u0932\u0902\u092c\u093e\u0908 = 5a + 4 \u0914\u0930 \u091a\u094c\u095c\u093e\u0908 = 5a + 3<\/p>\n

    \u0939\u0932 :\u00a0(ii) \u0915\u094d\u0937\u0947\u0924\u094d\u0930\u092b\u0932 : 35y2<\/sup>\u00a0+ 13y – 12
    \n\u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n35y2<\/sup>\u00a0+ 13y – 12\u00a0\u200b= 35y2<\/sup>\u200b + 28y – 15y – 12
    \n= 7y(5y + 4) – 3(5y + 4)
    \n= (5y + 4) (7y – 3)
    \n\u0905\u0924: \u0932\u0902\u092c\u093e\u0908 = 5y\u00a0+ 4 \u0914\u0930 \u091a\u094c\u095c\u093e\u0908 = 7y – 3<\/p>\n

    Ex 2.5 Class 9 \u0917\u0923\u093f\u0924\u00a0Q16. \u0918\u0928\u093e\u092d\u094b\u0902 (cuboids), \u091c\u093f\u0928\u0915\u0947 \u0906\u092f\u0924\u0928 \u0928\u0940\u091a\u0947 \u0926\u093f\u090f \u0917\u090f \u0939\u0948\u0902 \u0915\u093f, \u0935\u093f\u092e\u093e\u0913\u0902 \u0915\u0947 \u0932\u093f\u090f \u0938\u0902\u092d\u0935 \u0935\u094d\u092f\u0902\u091c\u0915 \u0915\u094d\u092f\u093e \u0939\u0948\u0902 ?\u00a0<\/strong>
    \n(i) \u0906\u092f\u0924\u0928 : 3x3<\/sup>\u00a0– 12x\u00a0<\/strong>
    \n(ii) \u0906\u092f\u0924\u0928 : 12ky2<\/sup>\u00a0+ 8ky – 20k\u00a0<\/strong>
    \n\u0939\u0932 :<\/strong>\u00a0(i) \u0906\u092f\u0924\u0928 : 3x3<\/sup>\u00a0– 12x
    \n\u0917\u0941\u0923\u0928\u0916\u0902\u0921 \u0915\u0930\u0928\u0947 \u092a\u0930
    \n\u0906\u092f\u0924\u0928 =\u00a03x3<\/sup>\u00a0– 12x = 3x(x – 4)
    \n\u091a\u0942\u0901\u0915\u093f \u0906\u092f\u0924\u0928 = L\u00a0\u00d7 B\u00a0\u00d7 H
    \n\u0905\u0924: L = 3, B = x \u0914\u0930 H = x – 4<\/p>\n

    \u0939\u0932 :<\/strong>\u00a0(ii) \u0906\u092f\u0924\u0928 : 12ky2<\/sup>\u00a0+ 8ky – 20k
    \n\u0906\u092f\u0924\u0928 =\u00a012ky2<\/sup>\u00a0+ 8ky – 20k
    \n= 4k (3y2<\/sup>\u00a0<\/sup>+ 2y – 5)
    \n= 4k (3y2<\/sup>\u00a0<\/sup>+ 5y – 3y – 5)
    \n= 4k [y (3y\u00a0<\/sup>+ 5)\u00a0– 1(3y + 5)]
    \n= 4k (3y\u00a0<\/sup>+ 5) (y – 1)
    \n\u091a\u0942\u0901\u0915\u093f \u0906\u092f\u0924\u0928 = L\u00a0\u00d7 B\u00a0\u00d7 H
    \n\u0905\u0924: L =\u00a04k, B =\u00a0(3y\u00a0<\/sup>+ 5)\u00a0\u0914\u0930 H =\u00a0(y – 1)<\/p>\n

    Hope given NCERT Solutions for Class 9 Maths Chapter 2 are helpful to complete your homework.<\/p>\n","protected":false},"excerpt":{"rendered":"

    NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (\u092c\u0939\u0941\u092a\u0926) (Hindi Medium) These Solutions are part of NCERT Solutions for Class 9 Maths in Hindi Medium. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials \u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.1\u00a0 Ex 2.1 Class 9 \u0917\u0923\u093f\u0924 Q1. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0935\u094d\u092f\u0902\u091c\u0915\u094b\u0902 \u092e\u0947\u0902 \u0915\u094c\u0928-\u0915\u094c\u0928 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 …<\/p>\n

    NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium)<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":7711,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[27,26,28,24,25],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium) - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-hindi\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium) - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (\u092c\u0939\u0941\u092a\u0926) (Hindi Medium) These Solutions are part of NCERT Solutions for Class 9 Maths in Hindi Medium. Here we have given NCERT Solutions for Class 9 Maths Chapter 2 Polynomials \u092a\u094d\u0930\u0936\u094d\u0928\u093e\u0935\u0932\u0940 2.1\u00a0 Ex 2.1 Class 9 \u0917\u0923\u093f\u0924 Q1. \u0928\u093f\u092e\u094d\u0928\u0932\u093f\u0916\u093f\u0924 \u0935\u094d\u092f\u0902\u091c\u0915\u094b\u0902 \u092e\u0947\u0902 \u0915\u094c\u0928-\u0915\u094c\u0928 \u090f\u0915 \u091a\u0930 \u092e\u0947\u0902 \u092c\u0939\u0941\u092a\u0926 … NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium) Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-hindi\/\" \/>\n<meta property=\"og:site_name\" content=\"NCERT Solutions\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\" \/>\n<meta property=\"article:published_time\" content=\"2021-07-23T11:27:57+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2021-07-23T12:33:38+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2019\/09\/NCERT-Solutions-for-Class-9-Maths-Chapter-2-Polynomials-Hindi-Medium.png\" \/>\n\t<meta property=\"og:image:width\" content=\"1280\" \/>\n\t<meta property=\"og:image:height\" content=\"720\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:site\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"14 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\",\"name\":\"NCERT Solutions\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\",\"https:\/\/www.instagram.com\/ncertsolutionsguru\/\",\"https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/\",\"https:\/\/in.pinterest.com\/ncertsolutionsguru\/\",\"https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training\",\"https:\/\/twitter.com\/ncertsolguru\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"width\":460,\"height\":45,\"caption\":\"NCERT Solutions\"},\"image\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/ncertsolutions.guru\/#website\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"name\":\"NCERT Solutions\",\"description\":\"NCERT Solutions for Class 1 to 12\",\"publisher\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/ncertsolutions.guru\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-hindi\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i2.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2019\/09\/NCERT-Solutions-for-Class-9-Maths-Chapter-2-Polynomials-Hindi-Medium.png?fit=1280%2C720&ssl=1\",\"contentUrl\":\"https:\/\/i2.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2019\/09\/NCERT-Solutions-for-Class-9-Maths-Chapter-2-Polynomials-Hindi-Medium.png?fit=1280%2C720&ssl=1\",\"width\":1280,\"height\":720,\"caption\":\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium)\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-hindi\/#webpage\",\"url\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-2-hindi\/\",\"name\":\"NCERT Solutions for Class 9 Maths Chapter 2 Polynomials (Hindi Medium) - 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