# Polynomials Class 10 Extra Questions Maths Chapter 2 with Solutions Answers

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
(d) More than 3

Question 2.
If 1 is zero of the polynomial p(x) = ax2 – 3(a – 1)x – 1, then the value of ‘a’ is:
(a) 1
(b) -1
(c) 2
(d) – 2
(a) 1

Question 3.
If a, ß are zeroes of x2 – 6x + k. What is the value of k if 3a + 2B = 20.
(a) – 16
(b) 8
(c) – 2
(d) -8
(a) – 16

Question 4.
If one zero of 2x2 – 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
(a) 2

Question 5.
If p(x) = x2 + 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
(b) 0

Question 6.
Dividing (x2 + 1) by (x + 1), the remainder will be:
(a) -1
(b) 11
(c) 0
(d) – 2
(c) 0

Question 7.
Dividing x3 + 3x + 3 by (x + 2), the remainder will be:
(a) -2
(b) -1
(c) 0
(d) 1
(d) 1

Question 8.
The zero’s of the polynomial (x2 – 2x – 3) will be:
(a) – 3, 1
(b) -3, -1
(c) 3, -1
(d) 3, 1
(c) 3, -1
Solution:
x2 – 2x – 3 = x2 – 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 x3 – 3x2 – x + 3 by x – 4x + 3 the remainder will be:
(a) – 3
(b) 3
(c) 1
(d) 0
(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 ax2 + bc + c and its zeroes be α and β.
(i) Here, α + β = $$\frac {1}{4}$$ and αβ = -1
Thus, the polynomial formed = x2 – (Sum of the zeroes)x + Product of the zeroes

If k = 4 then the polynomial is 4x2 – 3 – 4.

(ii) Here, α + β = √2 and αβ = $$\frac {1}{3}$$
Thus, the polynomial formed = x2 – (Sum of the zeroes) x + Product of the zeroes

If k = 3, then, the polynomial is 3x2 – 3√2x + 1.

(iii) Here, α + β = 0 and aß = √5
Thus, the polynomial formed = x2 – (Sum of the zeroes) x + Product of zeroes
= x2 – (0)x + √5
= x2 + √5

(iv) Let the polynomial be ax2 + 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 x2 – x + 1.

(v) Let the polynomial be ax2 + 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 4x2 + x + 1.

(vi) Let the polynomial be ax2 + 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 x2 – 4x + 1.

Question 2.
Divide p(y) by g(y) if p(y) = y3 – 3y2 – y + 3 and g(y) = y2 – 4y + 3.
We write
y3 – 3y2 – y + 3 = (y + 1) (y2 – 4y + 3).
Solution:

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

Question 3.
Examine if x – 1 is a factor of 2x3 – 5x + 3.
Solution:

Here, remainder is 0, hence, x – 1 is one factor of 2x3 – 5x + 3.

### Polynomials Class 10 Extra Questions Short Answer Type

Question 1.
Divide the polynomial f(x) = 3x2 – x3 – 3x + 5 by the polynomial g(x) = x – 1 – x2 and verify the division alogrithm.
Solution:
f(x) = – x3 + 3x2 – 3x + 5 and g(x) = – x2 + x- 1

∴ Quotient = x – 2 and remainder = 3.
as dividend = Q x divisior + remainder
– x3 + 3x2 – 3x + 5 = (x – 2)(-x2 + x – 1) +3
= – x3 + x2 – x + 2x2 – 2x + 2 + 3
= – x3 + 3x2 – 3x + 5
Verify

Question 2.
Divide 3 – x + 2x2 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
= 2x2 – 4x + 3x – 6 + 9
= 2x2 – x + 3
= dividend
Thus, (quotient x divisor) + remainder = dividend. Hence, the division algorithm is verified.

Question 3.
On dividing x3 – 3x2 + 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) = x2 – 3x2 + 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,
x3 + 3x2 + x + 2 = (x – 2) × g(x) + (-2x + 4)
⇒ x3 – 3x2 + x + 2 + 2x – 4 = (x – 2) × g(x)

On dividing x3 – 3x2 + 3x – 2 by x – 2, we get

Hence, g(x) = x2 – x + 1.

Question 4.
Divide 6x5 + 5x4 + 11x3 – 5x2 + 2x + 7 by 3x2 – 2x + 4.
Solution:

Question 5.
If – 3 is one of the zeros of the quadratic polynomial (k – 1) x2 + kx + 1. Find the value of other zeros.
Solution:
Given that – 3 is one zeros of the polynomial
(k – 1) x2 + 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) x2 + $$\frac {4}{3}$$ x + 1

⇒ x2 + 4x + 3 = x2 + 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) = 3x2 + 2x + 6,
degree of g(x) = 2
p(x) = 12x2 + 8x + 24,
degree of p(x) = 2

(i) Using division algorithm,
We have,’ p(x) = (x) × g(x) + r(x)
On dividing 12x2 + 8x + 24 by 3x2 + 2x + 6.
we get

Since, the remainder is zero, therefore 3x2 + 2x + 6 is a factor of 12x2 + 8x + 24.
∴ g(x) = 4 and r(x) = 0.

(ii) p(x) = x5 + 2x4 + 3x3 + 5x2 + 2
q(x) = x2 + x + 1, degree of g(x) = 2
g(x) = x3 + x2 + x + 1
r(x) = 2x2 – 2x + 1, degree of r(x) = 2
Here, deg q(x) = deg r(x)
On dividing x5 + 2x4 + 3x3 + 5x2 + 2 by x2 + x + 1, we get

Here, g(x) = x3 + x2 + x + 1
and r(x) = 2x2 – 2x + 1

(iii) Let p(x) = 2x4 + 8x3 + 6x2 + 4x + 12, r(x) = 2, degree of r(x) = 0
g(x) = x4 + 4x3 + 3x2 + 2x + 1
⇒ 9(x) = 10
Here deg r(x) = 0
On dividing 2x4 + 8x3 + 6x2 + 4x + 12 by 2,
we get

Question 7.
Find the zero’s of quadratic polynomial f(x) = 3x2 – 3 – 4. Verify the relationship between the zeros and its coefficients.
Solution:
∴ f(x) = 3x2 – x – 4
= 3x2 – 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 2x4 – 3x2 – 3x2 + 6x – 2. If two of its zeros are √2 and √2.
Solution.
quad. polynomial form by the given zeros is
(X – √2) (x + √2)
⇒ x2 – 2
Now divided the given polynomial by x2 – 2

∴ x4 – 3x3 – 3x2 + 6x – 2
= (x2 – 2) (2x2 – 3x + 1)
= (x2 – 2) (2x2 – 2x – x + 1)
= (x2 – 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

error: Content is protected !!