NCERT Class 9 Maths Chapter 2 Polynomials Exercise 2.3

Chapter 2 Polynomials Exercise 2.3

Question 1: Find the remainder when x³ + 3x² + 3x + 1 is divided by
i) x + 1
ii) x - 1/2
iii) x
iv) x + π
v) 5 + 2x
Answer:
Let p(x) be x³ + 3x² + 3x + 1
i) x + 1
x + 1 = 0
= x = -1

Therefore,
x³ + 3x² + 3x + 1
= (-1)³ + 3(-1)² + 3(-1) + 1
= -1 + 3 - 3 + 1
= 0

ii) x - 1/2
x - 1/2 = 0
= x = 1/2

Therefore,
x³ + 3x² + 3x + 1
= (1/2)³ + 3(1/2)² + 3(1/2) + 1
= 1/8 + 3/4 + 3/2 + 1
= 1/8 + 6/8 + 12/8 + 8/8
= (1 + 6 + 12 + 8)/8
= 27/8

iii) x
x = 0

Therefore,
x³ + 3x² + 3x + 1
= (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1
= 1

iv) x + π
x + π = 0
= x = -π

Therefore,
x³ + 3x² + 3x + 1
= (-π)³ + 3(-π)² + 3(-π) + 1
= -π³ + 3π² - 3π + 1

v) 5 + 2x
5 + 2x = 0
= 2x = -5
= x = -5/2

Therefore,
x³ + 3x² + 3x + 1
= (-5/2)³ + 3(-5/2)² + 3(-5/2) + 1
= -125/8 + 75/4 - 15/2 + 1
= -125/8 + 150/8 - 60/8 + 8/8
= (-125 + 150 - 60 + 8)/8
= -27/8

Question 2: Find the remainder when x³ - ax² + 6x - a is divided by x - a.
Answer:
Let p(x) be x³ - ax² + 6x - a

x - a = 0
= x = a

Therefore,
x³ - ax² + 6x - a
= (a)³ - a(a)² + 6(a) - a
= a³ - a³ + 6a - a
= 6a - a
= 5a

Hence the remainder when x³ - ax² + 6x - a is divided by x - a is 5a.

Question 3: Check whether 7 + 3x is a factor of 3x³ + 7x.
Answer:
7 + 3x will be the factor of 3x³ + 7x only when it divides 3x³ + 7x completely without leaving remainder.

Let p(x) be 3x³ + 7x

Now,
7 + 3x = 0
= 3x = -7
= x = -7/3

Therefore,
3x³ + 7x
= 3(-7/3)³ + 7(-7/3)
= 3(-343/27) + -49/3
= -343/9 - 49/3
= -343/9 - 147/9
= (-343 - 147)/9
= -490/9

Since -490/9 ≠ 0; therefore 7 + 3x is not a factor of 3x³ + 7x.