Bihar Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.2 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions 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) = r
^{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. Here, dividend and divisor are both in standard forms. So, we have:

2. Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as

x^{2} – x + 1.

We have:

∴ The quotient is x^{2} + x – 3 and the remainder is 8.

3. To carry out the division, we first write divisor in the standard form.

So, divisor = – x^{2} + 2

We have:

∴ The quotient is – x^{2} – 2 and the remainder is – 5x + 10.

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:

1. Let us divide 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 by t^{2} – 3.

We have:

Since the remainder is 0, therefore t^{2} – 3 is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

2. Let us divide 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 by x^{2} + 3x + 1.

We get,

Since, the remainder is 0, therefore x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

3. Let us divide x^{5} – 4x^{3} + x^{2} + 3x + 1 by x^{3} – 3x + 1. We get,

Here, remainder is 2(≠ 0). Therefore, x^{3} – 3x + 1 is not a factor of

x^{5} – 4x^{3} + x^{2} + 3x + 1.

Question 3.

Obtain all the 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:

Since two zeroes are \(\sqrt{\frac{5}{3}}\) and – \(\sqrt{\frac{5}{3}}\), so (x – \(\sqrt{\frac{5}{3}}\)) and (x + \(\sqrt{\frac{5}{3}}\)) are the factors of the given polynomial.

Now, (x – \(\sqrt{\frac{5}{3}}\)) (x + \(\sqrt{\frac{5}{3}}\)) = x^{2} – \(\frac{5}{3}\).

So, (3x^{2} – 5) is a factor of the given polynomial.

Applying the division algorithm to the given polynomial and 3x^{2} – 5, we have:

∴ 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 = (3x^{2} – 5)(x^{2} + 2x + 1)

Now, x^{2} + 2x + 1 = x^{2} + x + x + 1

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

= (x + 1)(x + 1)

So, its other zeroes are – 1 and – 1.

Thus, all the zeroes of the given fourth degree polynomial are \(\sqrt{\frac{5}{3}}\), – \(\sqrt{\frac{5}{3}}\), – 1 and – 1.

Question 4.

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

Solution:

Since on dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quoitent and remainder were (x – 2) and (- 2x + 4) respectively, therefore

Quoitent × Divisor + Remainder = Dividend

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

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

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

Let us divide x^{3} – 3x^{2} + 3x – 2 by x – 2. We get

∴ (1) gives 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:

There can be several examples for each of (i), (ii) and (iii).

However, one example for each case may be taken as under:

- p(x) = 14, g(x)= 2, q(x) = x
^{2}– x + 7, r(x) = 0 - p(x) = x
^{3}+ x^{2}+ x + 1, g(x) = x^{2}– 1, q(x) = x + 1, r(x) = 2x + 2 - p(x) = x
^{3}+ 2x^{2}– x + 2, g(x) = – 1, q(x) = x + 2, r(x) = 4