## 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:

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:

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:

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.

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