Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.3

Question 1.

Apply the division algorithm to find the quotient and remainder on dividing p (x) by g (x) as given below:

(i) p (x) = x^{3} – 3x^{2} + 5x – 3, g (x) = x^{2} – 2 (Pb. 2018 Set I, II, III)

(ii) p (x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

(iii) p (x) = x^{4} – 5x + 6, g (x) = 2 – x^{2} [Pb. 2017 Set-B]

Solution:

(i) Given that p (x) = x^{3} – 3x^{2} + 5x – 3 and g (x) = x^{2} – 2,

By division algorithm,

x^{3} – 3x^{2} + 5x – 3 = (x – 3) (x^{2} – 2) + (7x – 9)

Hence, quotient = x – 3 and remainder = 7x – 9

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

or p (x) = x^{4} + 0x^{3} – 3x^{2} + 4x + 5

and g (x) = x^{2} + 1 – x

or g (x) = x^{2} – x + 1

By Division Algorithm,

x^{4} – 3x^{2} + 4x + 5 = (x^{2} + x – 3) (x^{2} – x + 1) + 8

Hence, Quotient = x^{2} + x – 3 and remainder = 8

(iii) Given that p (x) = x^{4} – 5x + 6

or p (x) = x^{4} + 0x^{3} + 0x^{2} – 5x + 6

and g (x) = 2 – x^{2}

or g (x) = – x^{2} + 2

By division algorithtm,

x^{4} – 5x + 6 = (- x^{2} – 2) (- x^{2} + 2) + (- 5x + 10)

Hence, quotient = -x^{2} – 2.

remainder = – 5x + 10

Question 2.

Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:

(i) t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

(ii) x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

Solution:

∵ remainder is zero

∵ By division algorithm,

t^{2} – 3 is factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

(ii)

∵ remainder is zero

∴ By division algorithm, x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(iii)

∵ remainder is not zero.

∴ By division algorithm, x^{3} – 3x + 1 is not a factor of x^{5} – 4x^{3} + x^{2} + 3x + 1.

Question 3.

Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and – \(\sqrt{\frac{5}{3}}\).

Solution:

Given that two zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\)

∴ (x – \(\sqrt{\frac{5}{3}}\)) [x – (-\(\sqrt{\frac{5}{3}}\))] are factors of given polynomial or

(x – \(\sqrt{\frac{5}{3}}\)) (x + \(\sqrt{\frac{5}{3}}\)) are factors of given polynomial.

or x^{2} – \(\frac{5}{3}\) is a factor of given polynomial.

Now, apply division algorithm to given polynomial and x^{2} – \(\frac{5}{3}\)

∴ 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

= (x^{2} – \(\frac{5}{3}\)) (3x^{2} + 6x + 3)

= (x^{2} – \(\frac{5}{3}\)) (3) [x^{2} + 2x + 1]

= 3 (x^{2} – \(\frac{5}{3}\)) [x^{2} + 2x + 1]

[S = 2, P = 1]

= 3 (x^{2} – \(\frac{5}{3}\)) [x(x + 1) + 1 (x + 1)]

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

Now, other zeroes of polynomials are given by

x + 1 = 0 ; x = -1 or

x + 1 = 0 ; x = -1

The zeroes of the given fourth degree polynomial are :

\(\sqrt{\frac{5}{3}}\), –\(\sqrt{\frac{5}{3}}\), -1, -1.

Question 4.

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

Solution.

Compare given data with division algorithm, we have

p(x)= g (x). q(x) + r (x) or

p(x) – r(x) = g(x).q(x)

or g(x) . q (x) = p(x) – r(x)

or g(x) = \(\frac{p(x)-r(x)}{q(x)}\)

By putting various values, we get:

g(x) = \(\frac{\left(x^{3}-3 x^{2}+x+2\right)-(-2 x+4)}{x-2}\)

g(x) = \(\frac{x^{3}-3 x^{2}+x+2+2 x-4}{x-2}\)

g(x) = \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\) …………….(1)

Now,

∴ \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\) = x^{2} – x + 1

From (1) and (2), we get:

g(x) = x^{2} – x + 1

Question 5.

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

(i) deg p (x) = deg q (x)

(ii) deg r (x) = 0

(iii) deg q (x) = deg r (x)

Solution:

(i) Let p(x) = 5x^{2} – 5x +10; g(x) = 5 q(x) = x^{2} – x + 2; r(x) = 0

∴ By division algorithm,

5x^{2} – 5x + 10 = 5(x^{2} – x + 2) + 0

or p(x) = g(x) q(x) + r(x)

Also, deg p(x) = deg q(x) = 2

(ii) Let p(x) = 7x^{3} – 42x + 53

g(x) = x^{3} – 6x + 7

q(x) = 7; r(x) = 4

∴ By division algorithm,

7x^{3} -42x + 53 = 7(x^{3} – 6x+ 7)+ 4

or p(x) = q(x) g(x) + r(x)

Also, deg q(x) = 0 = deg r(x)

(iii) Let p(x) = 4x^{3} + x^{2} + 3x + 6;

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

q(x) = 4x – 11;

r(x) = 32x + 17

∴ By division algorithm,

4x^{3} + x^{2} + 3x + 6 = (4x – 11) (x^{2} + 3x + 1) + (32x + 17)

or p(x) = q(x) . g(x) + r(x)

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