## Polynomials

## Chapter 2

## Exercise 2.3

**EX 2.3 QUESTION 1.**

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

**(i) x + 1**

**(ii) x – 1/2**

**(iii) x**

**(iv) x + π**

**(v) 5 + 2x**

**Solution:**

(i) x+1= 0

⇒x = −1

p(−1) = (−1)^{3}+3(−1)^{2}+3(−1)+1

= −1+3−3+1

= 0

∴Remainder = 0

(ii) x- 1/2

x-1/2 = 0

⇒ x = 1/2

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

= (1/8)+(3/4)+(3/2)+1

= 27/8

∴Remainder = 27/8

(iii) x = 0

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

= 1

∴Remainder = 0

(iv)

x+π = 0

⇒ x = −π

p(0) = (−π)^{3 }+3(−π)^{2}+3(−π)+1

= −π^{3}+3π^{2}−3π+1

∴Remainder = −π^{3}+3π^{2}−3π+1

5+2x=0

⇒ 2x = −5

⇒ x = -5/2

(-5/2)^{3}+3(-5/2)^{2}+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1

= -27/8

∴Remainder =-27/8

**EX 2.3 QUESTION 2.**

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

**Solution:**

Let p(x) = x^{3}−ax^{2}+6x−a

x−a = 0

∴x = a

p(a) = (a)^{3}−a(a^{2})+6(a)−a

= a^{3}−a^{3}+6a−a = 5a

∴ Remainder = a^{3}−a^{3}+6a−a = 5a

**EX 2.3 QUESTION 3.**

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

Solution:

7+3x = 0

⇒ 3x = −7

⇒ x = -7/3

3(-7/3)^{3}+7(-7/3)

= -(343/9)+(-49/3)

= (-343-(49)3)/9

= (-343-147)/9

= -490/9 ≠ 0

∴Remainder ≠ 0

∴7+3x is not a factor of 3x^{3}+7x