Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.3

Question 1.

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

(i) x + 1

Answer:

By remainder theorem, the required remainder is equal to p(- 1).

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

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

= -1 + 3 – 3 + 1 = 0

Hence, required remainder = p(- 1) = 0.

(ii) x – \(\frac{1}{2}\)

Answer:

By remainder theorem, the required remainder is equal to p(\(\frac{1}{2}\))

Now, p(x) = x^{3} + 3x^{3} + 3x + 1

∴ p\(\left(\frac{1}{2}\right)\) = \(\left(\frac{1}{2}\right)\)^{3} + 3\(\left(\frac{1}{2}\right)\)^{2} + 3\(\left(\frac{1}{2}\right)\) + 1

= \(\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1=\frac{1+6+12+8}{8}=\frac{27}{8}\)

(iii) x

Answer:

By remainder theorem, the required remainder is equal to p(0).

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

p( 0) = 0 + 0 + 0 + 1 = 1

Hence, the required remainder = p(0) = 1.

(iv) x + π

Answer:

By remainder theorem, the required remainder is p(- n).

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

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

= – π^{3} + 3π^{2} – 3π + 1

(v) 5 + 2x

Answer:

By remainder theorem, the required remainder is p(- \(\frac{5}{2}\))

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

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

= \(-\frac{125}{8}+\frac{75}{4}-\frac{15}{2}+1=\frac{-125+150-60+8}{8}=\frac{-27}{8}\)

Question 2.

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

Answer:

Let p(x) = x^{3} – ax^{2} + 6x – a

By remainder theorem, when p(x) is divided by x – a, then remainder = p(a)

∴ p(a) = a^{3} – a. a^{2} + 6a – a

= a^{3} – a^{3} + 6a – a = 5a

Question 3.

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

Answer:

7 + 3x will be a factor of p(x) = 3x^{3} + 7x, if p\(\) = 0

Now, p\(\left(-\frac{7}{3}\right)\) = 3\(\left(-\frac{7}{3}\right)\)^{3} + 7\(\left(-\frac{7}{3}\right)\) = 3\(\frac{-343}{27}-\frac{49}{3}=-\frac{343}{9}-\frac{49}{3}=\frac{-490}{9}\) ≠ 0

∴ 7 + 3x is not a factor of 3x^{3} + 7x.

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