RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3

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 x³ + 3x² + 3x + 1 is divided by :
(i) x + 1
Answer:
By remainder theorem, the required remainder is equal to p(- 1).
Now, p(x) = x³ + 3x² + 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) = x³ + 3x³ + 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) = x³ + 3x² + 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) = x³ + 3x² + 3x + 1
p(- π) = (- π)³ + 3(- π)² + 3(- π) + 1
= – π³ + 3π² – 3π + 1

(v) 5 + 2x
Answer:
By remainder theorem, the required remainder is p(- \(\frac{5}{2}\))
Now, p(x) = x³ + 3x² + 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 x³ – ax² + 6x – a is divided by x – a.
Answer:
Let p(x) = x³ – ax² + 6x – a
By remainder theorem, when p(x) is divided by x – a, then remainder = p(a)
∴ p(a) = a³ – a. a² + 6a – a
= a³ – a³ + 6a – a = 5a

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
7 + 3x will be a factor of p(x) = 3x³ + 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 3x³ + 7x.