Gujarat Board GSEB Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.3

Question 1.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

- p(x) = x
^{3}– 3x^{2}+ 5x – 3; g(x) = x^{2}– 2 - p(x) = x
^{4}– 3x^{2}+ 4x + 5, g(x) = x^{2}+ 1 – x - p(x) = x
^{4}– 5x + 6, g(x) = 2 – x^{2}

Solution:

1. By applying the division algorithm to the given polynomials p(x) and g(x)

âˆ´ Remainder = 7x – 9

degree of 7x – 9 = 1 < degree of (x^{2} – 2) = 2.

â‡’ Quotient = x – 3

Remainder = 7x – 9

Verify: Quotient Ã— Division + Remainder

= (x – 3) (x^{2} – 2) + (7x – 9)

= x^{3} – 2x – 3x^{2} + 6 + 7x – 9

= x^{3} – 3x^{2} + 5x – 3 = Dividend.

2. p(x) = x^{4} – 3x^{2} + 4x + 5

g(x) = x^{2} + 1 – x

Standard form of g(x) = x^{2} – x + 1.

Now, by applying the division algorithm to the given polynomials p(x) and g(x)

degree of 8 = 0 < degree of x^{2} – x + 1 = 2

So, Quotient = x^{2} + x – 3

Remainder = 8.

Verify: Quotient Ã— Divisor + Remainder

= (x^{2} + x – 3) (x^{2} – x + 1) + 8

= x^{4} – x^{3} + x^{2} + x^{3} – x^{2} + x – 3x^{2} + 3x – 3 + 8

= x^{4} – 3x^{2} + 4x + 5

= Dividend

3. p(x) = x^{4} – 5x + 6

g(x) = 2 – x^{2}

Standard form of g(x) = -x^{2} + 2

Now, by applying division algorithm to the given polynomial p(x) and g(x).

âˆµ deg. of -5x + 10 = 1 < deg. of -x^{2} + 2 = 2

So, Quotient = -x^{2} – 2

Remainder = -5x + 10.

Verify: Quotient Ã— Divisor + Remainder

= (-x^{2} – 2) Ã— (-x^{2} + 2) + (-5x + 10)

= (-x^{2})^{2} – 4 – 5x + 10

= x^{4} – 5x + 6 = Dividend

Question 2.

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

- t
^{2}– 3, 2t^{4}+ 3t^{3}– 2t^{2}– 9t – 12 - x
^{2}+ 3x + 1, 3x^{4}+ 5x^{3}– 7x^{2}+ 2x + 2 - x
^{3}– 3x + 1, x^{5}– 4x^{3}+ x^{2}+ 3x + 1

Solution:

By applying division algorithm to the given polynomials,

Since, the remainder is 0, hence the first polynomial is a factor of the second polynomial.

2. x^{2} + 3x + 1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

By applying division algorithm to the given polynomials,

Since the remainder is 0, hence the first polynomial is a factor of the second polynomial.

3. x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x + 1

By applying division algorithm to the given polynomials,

Since the remainder is 2 (â‰ 0), hence the first polynomial is not a factor of second polynomial.

Question 3.

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

Solution:

Since two zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\) hence

\(\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=x^{2}-\frac{5}{3}\) is a factor of the given polynomial.

Now, we divide the given polynomial by x^{2} – \(\frac{5}{3}\).

So, 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5

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

Now, 3x^{2} + 6x + 3

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

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

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

So, its zeroes are -1 and -1.

â‡’ Remaining two zeroes of the given polynomial are -1 and -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:

p(x) = x^{3} – 3x^{2} + x + 2

q(x) = x – 2

r(x) = -2x + 4

By division algorithm for polynomials

p(x) = g(x) Ã— q(x) + r(x)

â‡’ x^{3} – 3x^{2} + x + 2

= (x – 2) g(x) + (-2x + 4)

â‡’ (x – 2) g(x)

= x^{3} – 3x^{2} + x + 2 + 2x – 4

= x^{3} – 3x^{2} + 3x – 2

âˆ´ g(x) = (x^{3} – 3x^{2} + 3x – 2) Ã· (x – 2)

âˆ´ g(x) = x^{2} – x + 1

Question 5.

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

- deg p(x) = deg q(x)
- deg q(x) = deg r(x)
- deg r(x) =0.

Solution:

1. deg p(x) = deg q(x)

Let p(x) = 8x^{2} – 8x + 16

and g(x) = 8

âˆ´ q(x) = x^{2} – x + 2

r(x) = 0

p(x) = g(x) Ã— q(x) + r(x)

2. deg q(x) = deg r(x)

Let p(x) = x^{3} + x^{2} + x + 1

and g(x) = x^{2} – 1

âˆ´ q(x) = x + 1

r(x) = 2x + 2

p(x) = g(x) Ã— q(x) + r(x)

3. deg r(x) = 0

Let p(x) = x^{4} + x^{3} + x^{2} – 2x – 3

and g(x) = x^{2} – 2

âˆ´ q(x) = x^{2} + x + 3

r(x) = 3

p(x) = g(x) Ã— q(x) + r(x)