Here you will find Polynomials Class 10 Extra Questions Maths Chapter 2 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Polynomials with Answers Solutions

**Extra Questions for Class 10 Maths Chapter 2 Polynomials with Solutions Answers**

### Polynomials Class 10 Extra Questions Objective Type

Question 1.

The number of polynomials having zeroes -2 and 5 is:

(a) 1

(b) 2

(c) 3

(d) More than 3

Answer:

(d) More than 3

Question 2.

If 1 is zero of the polynomial p(x) = ax^{2} – 3(a – 1)x – 1, then the value of ‘a’ is:

(a) 1

(b) -1

(c) 2

(d) – 2

Answer:

(a) 1

Question 3.

If a, ß are zeroes of x^{2} – 6x + k. What is the value of k if 3a + 2B = 20.

(a) – 16

(b) 8

(c) – 2

(d) -8

Answer:

(a) – 16

Question 4.

If one zero of 2x^{2} – 3x + k is reciprocal to the other, then the value of k is:

(a) 2

(b) \(\frac {-2}{3}\)

(c) \(\frac {-3}{2}\)

(d) -3

Answer:

(a) 2

Question 5.

If p(x) = x^{2} + 6x + 9 and q(x) = x + 3 then remainder will be when p(x) is divided by q(x):

(a) – 1

(b) 0

(c) 11

(d) 2

Answer:

(b) 0

Question 6.

Dividing (x^{2} + 1) by (x + 1), the remainder will be:

(a) -1

(b) 11

(c) 0

(d) – 2

Answer:

(c) 0

Question 7.

Dividing x^{3} + 3x + 3 by (x + 2), the remainder will be:

(a) -2

(b) -1

(c) 0

(d) 1

Answer:

(d) 1

Question 8.

The zero’s of the polynomial (x^{2} – 2x – 3) will be:

(a) – 3, 1

(b) -3, -1

(c) 3, -1

(d) 3, 1

Answer:

(c) 3, -1

Solution:

x^{2} – 2x – 3 = x^{2} – 3x + x – 3

= x(x – 3) + 1 (x – 3 ) = (x – 3) (x + 1)

∴ Zeros of the polynomial are 3, -1

Hence, choice (c) is correct.

Question 9.

Dividing x^{3} – 3x^{2} – x + 3 by x – 4x + 3 the remainder will be:

(a) – 3

(b) 3

(c) 1

(d) 0

Answer:

(d) 0

### Polynomials Class 10 Extra Questions Very Short Answer Type

Question 1.

Find a quadratic polynomial each with the given numbers as the sum and product of the zeroes respectively.

(i) \(\frac {1}{4}\), -1

(ii) √2, \(\frac {1}{3}\)

(iii) 0, √5

(iv) 1, 1

(v) – \(\frac {1}{4}\), \(\frac {1}{4}\)

(vi) 4, 1

Solution:

Let the polynomial be ax^{2} + bc + c and its zeroes be α and β.

(i) Here, α + β = \(\frac {1}{4}\) and αβ = -1

Thus, the polynomial formed = x^{2} – (Sum of the zeroes)x + Product of the zeroes

If k = 4 then the polynomial is 4x^{2} – 3 – 4.

(ii) Here, α + β = √2 and αβ = \(\frac {1}{3}\)

Thus, the polynomial formed = x^{2} – (Sum of the zeroes) x + Product of the zeroes

If k = 3, then, the polynomial is 3x^{2} – 3√2x + 1.

(iii) Here, α + β = 0 and aß = √5

Thus, the polynomial formed = x^{2} – (Sum of the zeroes) x + Product of zeroes

= x^{2} – (0)x + √5

= x^{2} + √5

(iv) Let the polynomial be ax^{2} + bx + c and its zeroes α and β. Then

If a = 1 then b = -1 and c = 1

∴ One quadratic polynomial which satisfy the given conditions is x^{2} – x + 1.

(v) Let the polynomial be ax^{2} + bx +c and its zeroes be a and B. Then,

if a = 4, then b = 1 and c = 1.

∴ One quadratic polynomial which satisfy the given conditions is 4x^{2} + x + 1.

(vi) Let the polynomial be ax^{2} + bx + c and its zeroes be a and ß. then,

and αβ = 1 if a = 1 then b = -4 and c = 1

∴ One quadratic polynomial which satisfy the given conditions is x^{2} – 4x + 1.

Question 2.

Divide p(y) by g(y) if p(y) = y^{3} – 3y^{2} – y + 3 and g(y) = y^{2} – 4y + 3.

Answer:

We write

y^{3} – 3y^{2} – y + 3 = (y + 1) (y^{2} – 4y + 3).

Solution:

Here, the quotient is y + 1 and the remainder is zero.

Question 3.

Examine if x – 1 is a factor of 2x^{3} – 5x + 3.

Solution:

Here, remainder is 0, hence, x – 1 is one factor of 2x^{3} – 5x + 3.

### Polynomials Class 10 Extra Questions Short Answer Type

Question 1.

Divide the polynomial f(x) = 3x^{2} – x^{3} – 3x + 5 by the polynomial g(x) = x – 1 – x^{2} and verify the division alogrithm.

Solution:

f(x) = – x^{3} + 3x^{2} – 3x + 5 and g(x) = – x^{2} + x- 1

∴ Quotient = x – 2 and remainder = 3.

as dividend = Q x divisior + remainder

– x^{3} + 3x^{2} – 3x + 5 = (x – 2)(-x^{2} + x – 1) +3

= – x^{3} + x^{2} – x + 2x^{2} – 2x + 2 + 3

= – x^{3} + 3x^{2} – 3x + 5

Verify

Question 2.

Divide 3 – x + 2x^{2} by (2 – x) and verify alogrithm.

Solution:

First we write the terms of dividend and divisor in decreasing order of their degrees and then perform the division as shown below:

Clearly, degree (9) = 0 < degree (- x + 2).

∴ quotient = (- 2x – 3) and remainder = 9.

= (quotient x divisor) + remainder

= (- 2x – 3)(- x + 2) + 9

= 2x^{2} – 4x + 3x – 6 + 9

= 2x^{2} – x + 3

= dividend

Thus, (quotient x divisor) + remainder = dividend. Hence, the division algorithm is verified.

Question 3.

On dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and – 2x + 4, respectvely. Find g(x).

Solution:

Given, p(x) = x^{2} – 3x^{2} + x + 2, q(x) = x – 2 and r(x) = – 2x + 4. By division algorithm, we know that

Dividend = Divisor × Quotient + Remainder

p(x) = q(x) × g(x) + r(x).

Therefore,

x^{3} + 3x^{2} + x + 2 = (x – 2) × g(x) + (-2x + 4)

⇒ x^{3} – 3x^{2} + x + 2 + 2x – 4 = (x – 2) × g(x)

⇒

On dividing x^{3} – 3x^{2} + 3x – 2 by x – 2, we get

Hence, g(x) = x^{2} – x + 1.

Question 4.

Divide 6x^{5} + 5x^{4} + 11x^{3} – 5x^{2} + 2x + 7 by 3x^{2} – 2x + 4.

Solution:

Question 5.

If – 3 is one of the zeros of the quadratic polynomial (k – 1) x^{2} + kx + 1. Find the value of other zeros.

Solution:

Given that – 3 is one zeros of the polynomial

(k – 1) x^{2} + kx + 1

∴ (k – 1)(-3)^{2} + k (-3) + 1 = 0 .

⇒ 9k – 9 – 3k + 1 = 0

⇒ 6k = 8

⇒ Putting the value of k = \(\frac {4}{3}\) in given polynomial,

we get (\(\frac {4}{3}\)– 1) x^{2} + \(\frac {4}{3}\) x + 1

⇒ x^{2} + 4x + 3 = x^{2} + 3x + x + 3

= x (x + 3) + 1 (x + 3)

⇒ (x + 3) (x + 1)

∴ zeros are – 3 and – 1

Hence their zeros of the quadratic polynomial is – 1.

Question 6.

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg g(x)

(ii) deg g(x) = deg r(x)

(iii) deg q(x) = 0

Solution:

Let

q(x) = 3x^{2} + 2x + 6,

degree of g(x) = 2

p(x) = 12x^{2} + 8x + 24,

degree of p(x) = 2

(i) Using division algorithm,

We have,’ p(x) = (x) × g(x) + r(x)

On dividing 12x^{2} + 8x + 24 by 3x^{2} + 2x + 6.

we get

Since, the remainder is zero, therefore 3x^{2} + 2x + 6 is a factor of 12x^{2} + 8x + 24.

∴ g(x) = 4 and r(x) = 0.

(ii) p(x) = x^{5} + 2x^{4} + 3x^{3} + 5x^{2} + 2

q(x) = x^{2} + x + 1, degree of g(x) = 2

g(x) = x^{3} + x^{2} + x + 1

r(x) = 2x^{2} – 2x + 1, degree of r(x) = 2

Here, deg q(x) = deg r(x)

On dividing x^{5} + 2x^{4} + 3x^{3} + 5x^{2} + 2 by x^{2} + x + 1, we get

Here, g(x) = x^{3} + x^{2} + x + 1

and r(x) = 2x^{2} – 2x + 1

(iii) Let p(x) = 2x^{4} + 8x^{3} + 6x^{2} + 4x + 12, r(x) = 2, degree of r(x) = 0

g(x) = x^{4} + 4x^{3} + 3x^{2} + 2x + 1

⇒ 9(x) = 10

Here deg r(x) = 0

On dividing 2x^{4} + 8x^{3} + 6x^{2} + 4x + 12 by 2,

we get

Question 7.

Find the zero’s of quadratic polynomial f(x) = 3x^{2} – 3 – 4. Verify the relationship between the zeros and its coefficients.

Solution:

∴ f(x) = 3x^{2} – x – 4

= 3x^{2} – 4x + 3x – 4

= x(3x – 4) + 1 (3x – 4)

= (3x – 4) (x + 1)

∴ zero’s are \(\frac {4}{3}\) and -1.

sum of the zeros =

Question 8.

Solve the pair of linear simultaneous equations

2x – y = 1 and x + 2y = 13

By drawing their graphs. Find the coordinates of vertices of a triangle formed by these lines and y-axis.

Solution:

2x – y = 1 ⇒ x + 2y = 13

2x = 1 + y ⇒ x = 13 – 2y

Now plot the points on the graph paper.

The coordinate of required ∆ are

A(3, 5), B (0, \(\frac {13}{2}\)) and C(0, -1) respectively

Solution is x = 3, y = 5

Question 9.

Find all the zeros of polynomial 2x^{4} – 3x^{2} – 3x^{2} + 6x – 2. If two of its zeros are √2 and √2.

Solution.

quad. polynomial form by the given zeros is

(X – √2) (x + √2)

⇒ x^{2} – 2

Now divided the given polynomial by x^{2} – 2

∴ x^{4} – 3x^{3} – 3x^{2} + 6x – 2

= (x^{2} – 2) (2x^{2} – 3x + 1)

= (x^{2} – 2) (2x^{2} – 2x – x + 1)

= (x^{2} – 2) [2x(x -1) – 1(x-1)]

= (x √2 )(x + √2)(x – 1) (2x – 1)

Hence all zeros are

1/2, 1, √2 and – √2