Jharkhand Board JAC Class 9 Maths Solutions Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.3

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Question 1.

Find the remainder when x^{3} + 3x^{2} + 3x + 1 is divided by

(i) x + 1

(ii) x – \(\frac{1}{2}\)

(iii) x

(iv) x + π

(v) 5 + 2x

Answer:

(i) x + 1

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

By remainder theorem, remainder = P(-1)

= (-1)^{3} + 3 (-1)^{2} + 3(-1) + 1

= -1 + 3 – 3 + 1

= 0

Therefore, the remainder is 0.

(ii) x – \(\frac{1}{2}\)

By remainder theorem,

Remainder = p(\(\frac{1}{2}\))

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

= \(\frac{1}{8}\) + \(\frac{3}{4}\) + \(\frac{3}{2}\) + 1

= \(\frac{27}{8}\)

Therefore, the remainder is \(\frac{27}{8}\).

(iii) x

By remainder theorem,

Remainder = p (0)

= 0^{3} + 3(0)^{2} + 3(0) + 1

= 1

Therefore, the remainder is 1.

(iv) x + π

By remainder theorem,

Remainder = p (-π)

= (-π)^{3} + 3 (-π)^{2} + 3 (-π) + 1

= -π^{3} + 3π^{2} – 3π + 1

Therefore, the remainder is -π^{3} + 3π^{2} – 3π + 1.

(v) 5 + 2x

By remainder theorem,

Remainder = p(\(– \frac{5}{2}\))

= \(\left(-\frac{5}{2}\right)^3+3\left(-\frac{5}{2}\right)^2+3\left(-\frac{5}{2}\right)+1\)

= \(\frac{-125+150-60+8}{8}\)

= \(– \frac{27}{8}\)

Therefore, the remainder is \(– \frac{27}{8}\).

Question 2.

Find the remainder when x^{3} – ax^{2} + 6x – a is divided by x – a.

Answer:

Lef f(x) = x^{3} – ax^{2} + 6x – a be divided by x – a

By remainder theorem,

Remainder = f(a)

= a^{3} – a(a)^{2} + 6(a) – a

= a^{3} – a^{3} + 6a – a

= 5a

Therefore, remainder obtained is 5a when x^{3} – ax^{2} + 6x – a is divided by x – a.

Question 3.

Check whether 7+3x is a factor of 3x^{3} + 7x.

Ans. Let f(x) = 3x^{2} + 7x be divided by 7 + 3x

As 7 + 3x = 0 gives

x = \(– \frac{7}{3}\)

By Remainder theorem,

Remainder = f(\(-\frac{7}{3}\))

= \(3\left(-\frac{7}{3}\right)^3+7\left(-\frac{7}{3}\right)\)

= \(-\frac{343}{9}-\frac{49}{3}\)

= \(\frac{-343-147}{9}\)

= \(-\frac{490}{9}\) ≠ 0

As remainder is not zero, so 7 + 3x is not a factor of 3x^{3} + 7x.