Jharkhand Board JAC Class 10 Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 2 Polynomials Exercise 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:

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

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

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

Solution:

1.

Quotient x – 3, Remainder = 7x – 9

2. p(x) = x^{4} – 3x^{2} + 4x + 5

= x^{4} + 0x^{3} – 3x^{2} + 4x + 5

and g(x) = x^{2} + 1 – x = x^{2} – x + 1

Quotient = x^{2} + x – 3, Remainder = 8

3. p(x) = x^{4} – 5x + 6

= x^{4} + 0x^{3} + 0x^{2} – 5x + 6

and g(x) = 2 – x^{2} = -x^{2} + 2

Quotient = -x^{2} – 2, Remainder = – 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:

1. t^{2} – 3, 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12

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

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

Solution:

1.

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

2.

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

3.

As the remainder is not 0, 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:

Since \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are given two zeroes of the polynomial, \(\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=x^2-\frac{5}{3}\) is a factor of the given polynomial. Then, to obtain the other zeroes of the polynomial, we divide it by x^{2} – \(\frac{5}{3}\)

Now,

3x^{2} + 6x + 3 = 3(x^{2} + 2x + 1)

= 3(x + 1)^{2}

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

Hence, the two zeroes of 3x^{2} + 6x + 3 are 1 and – 1.

Hence, the two zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 other than the given zeroes 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:

Here, dividend p(x) = x^{3} – 3x^{2} + x + 2, quotient q(x) = (x – 2) and remainder r(x) = (-2x + 4).

Now, p(x) = g(x) × q(x) + r(x)

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

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

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

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

Hence, 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

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

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

3. deg r(x) = 0.

Solution:

1. deg p (x) = deg q (x) implies that deg g (x) = 0. i.e., g(x) is a non-zero constant. One such example can be given as p (x) = 3x^{2} + 15x + 13, g(x) = 3, q(x) = x^{2} + 5x + 4 and r(x) = 1, which satisfies the division algorithm as:

3x^{2} + 15x + 13 = 3(x^{2} + 5x + 4) + 1.

2. deg q(x) = deg r (x) implies that deg g (x) > deg q(x), because

deg g (x) > degr (x). One such example can be given as p (x) = x^{3} + 5x^{2} + 2x + 7, g(x) = x^{2} + 1, q(x) = x + 5 and r(x) = x + 2, which satisfies the division algorithm as:

x^{3} + 5x^{2} + 2x + 7 = (x^{2} + 1)(x + 5) + (x + 2).

3. deg r(x) = 0 implies that the remainder is a constant. One such example can be given as p (x) = x^{3} + 4x^{2} + 5x + 9, g(x) = x + 3, q(x) = x^{2} + x + 2 and r(x) = 3, which satisfies the division algorithm as:

x^{3} + 4x^{2} + 5x + 9 = (x + 3)(x^{2} + x + 2) + 3.

Note: There can be several examples in each of the above cases.