These NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.4
Question 1.
Multiply the binomials.
(i) (2x + 5) and (4x – 3)
(ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m)
(iv) (a+ 3b) and (x + 5)
(v) (2pq + 3q2) and (3pq -2q2)
(vi) \(\left(\frac{3}{4} a^{2}+3 b^{2}\right)\) and \(\left(a^{2}-\frac{2}{3} b^{2}\right)\)
Answer:
(2x + 5) (4x – 3) = 2x (4x – 3) + 5 (4x – 3)
= 2x × 4x + 2x (-3) + 5 × 4x + 5 (- 3)
= 8x2 – 6x + 20x – 15
= 8x2 + 14x – 15
(ii) (y – 8) (3y – 4) = y (3y – 4) – 8 (3y – 4)
= y × 3y + y (-4) – [8 × 3y + 8 (-4)]
= 3y2 – 4y – (24y – 32)
= 3y2 – 4y – 24y + 32
= 3y2 – 28y + 32
(iii) (2.5l – 0.5m) (2.5l + 0.5m)
= 2.5l (2.5l + 0.5m) – 0.5m (2.5l + 0.5m)
= 2.5l × 2.5l + 2.5l × 0.5m – [0.5m × 2.5l + 0.5m × 0.5m]
= 6.25l2 + 1.25lm – 1.25lm – 0.25m2
= 6.25l2 – 0.25m2
(iv) (a + 3b) (x + 5) = a × (x + 5) + 3b × (x + 5)
= ax + 5a + 3bx + 15b
= ax + 5a + 3bx + 15b
(v) (2pq + 3q2) × (3pq – 2q2)
= 2pq × (3pq – 2q2) + 3q2 × (3pq – 2q2)
= 6p2q2 – 4pq3 + 9pq3 – 6q4
= 6p2q2 + 5pq3 – 6q4
= 3a4 -2a2b2 +12a2b2 – 8b4
= 3a4 + 10a2b2 – 8b4
Question 2.
Find the product.
(i) (5 – 2x) (3 + x)
(ii) (x + 7y) (7x – y)
(iii) (a2 + b) (a + b2)
(iv) (p2 – q2) (2p + q)
Answer:
(i) (5 – 2x) (3 + x) = 5 × (3 + x) – 2x (3 + x)
= 5 × 3 + 5 × x – (2x) × 3 – 2x (x)
= 15 + 5x – 6x – 2x2 = 15 – x – 2x2
(ii) (x + 7y)(7x – y) = x × (7x – y) + 7y × (7x – y)
= x × 7x – x × y + 7y × 7x – 7y × y
= 7x2 – xy + 49xy – 7y2
= 7x2 + 48xy – 7y2
(iii) (a2 + b)(a + b2) = a2 × (a + b2) + b × (a + b2)
= a2 × a + a2 × b2 + b × a + b × b2
= a3</sup + a2b2 + ab + b3
(iv) (p2 – q2) (2p + q)
= p2 × (2p + q) – q2 (2p + q)
= p2 × 2p + p2 × q – q2 × 2p – q2 × q
= 2p3 + p2q – 2pq2 – q3
Question 3.
Simplify
(i) (x2 – 5) (x + 5) + 25
(ii) (a2 + 5) (b3 + 3) + 5
(iii) (t + s2) (t2 – s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)
(v) (x + y) (2x + y) + (x + 2y) (x – y)
(vi) (x + y) (x2 – xy + y2)
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 45x + 12y
(viii) (a + b + c) (a + b – c)
Answer:
(i) (x2 – 5) (x + 5) + 25
= x2 × (x + 5) – 5 (x + 5) + 25
= x2× x + x2 × 5 – 5 × x – 5 × 5 + 25
= x3 + 5x2 – 5x – 25 + 25
= x3 + 5x2 – 5x
(ii) (a2 + 5) (b3 + 3) + 5
= a2 × (b3 + 3) + 5 × (b3 + 3) + 5
= a2b3 + 3a2 + 5b3 + 15 + 5
= a2b3 + 3a2 + 53 + 20
(iii) (t + s2) (t2 – s) = t × (t2 – s) + s2 × (t2 – s)
= t × t2 – t × s + s2 × t2 – S2 × s
= t3 – ts + s2t2 – s3
(iv) (a + b)(c – d) + (a – b)(c + d) + 2(ac + bd)
= a × (c – d) + b × (c – d) + a × (c + d) – b × (c + d ) + 2 × ac + 2 × bd
= a × c – a × d+ b × c – b × d + a × c + a × d – b × c – b × d + 2ac + 2bd
= ac – ad + bc – bd + ac + ad – bc – bd + 2ac + 2bd
= ac + ac + 2ac – ad + ad + bc – bc – bd – bd + 2bd
= 4ac – 2bd + 2bd = 4ac
(v) (x + y) (2x + y) + (x + 2y) (x – y)
= x × (2x + y) + y (2x + y) + x × (x – y) + 2y × (x – y)
= x × 2x + x × y + y × 2x + y × y + x × x + x × (-y) + 2y × x + 2y × (-y)
= 2x2 + xy + 2xy + y2 + x2 – xy + 2xy – 2y2 = 2x2 + x2 + xy + 2xy + 2xy – xy + y2 – 2y2
= 3x2 + 5xy – xy – y2
= 3x2 + 4xy – y2
(vi) (x + y) (x2 – xy + y2)
= x × (x2 – xy + y2) + y × (x2 – xy + y2)
= x × x2 + x × (-xy) + x × y2 + y × x2 + y × (-xy) + y × y2
= x3 – x2y + xy2 + x2y – xy2 + y3
= x3 – x2y + x2y + xy2 – xy2 + y3
= x3 + y3
(vii) (1.5x – 4y) (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x × (1.5x + 4y + 3) – 4y × (1.5x + 4y + 3) – 4.5x + 12y
= 1.5x × 1.5x + 1.5x × 4y + 1.5x × 3 – 4y × 1.5x – 4y × 4y – 4y × 3 – 4.5x + 12y
= 2.25xz + 6xy + 4.5x – 6xy – 16y2
– 12y – 4.5x + 12y
= 2.25x2 + 6xy – 6xy + 4.5x – 4.5x – 16y2 – 12y + 12y
= 2.25x2 + 0 + 0 – 16y2 + 0
= 2.25x2 – 16y2
(viii) (a + b + c) (a + b – c)
= a × (a + b – c) + b × (a + b – c) + c × (a + b – c)
= a × a + a × b + a × -c + b × a + b × b – b × c + c × a + b × c – c × c
= a2 + ab – ac + ab + b2 – bc + ac + bc – c2
= a2 + ab + ab – ac + ac – bc + bc + b2 – c2
= a2 + 2ab + o + o + b2 – c2
= a2 + b2 – c2 + 2ab