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

^{3 }+ 3x

^{2 }+ 3x + 1 is divided by

**(i) x + 1**

Solution:x + 1 = 0

x = −1

Therefore remainder will be f(x):

f(−1) = (−1)

^{3 }+ 3(−1)^{2 }+ 3(−1) + 1= −1 + 3 − 3 + 1

= 0

**(ii) x – 1/2**

Solution:x – 1/2 = 0

x = 1/2

Therefore remainder will be f(x):

f(1/2) = (1/2)

^{3}+ 3(1/2)^{2}+ 3(1/2) + 1= (1/8) + (3/4) + (3/2) + 1

= 27/8

**(iii) x**

Solution:x = 0

Therefore remainder will be f(x):

f(0) = (0)

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

**(iv) x + pi**

Solution:x + pi = 0

x = −pi

Therefore remainder will be f(x):

f(−pi) = (−pi)

^{3 }+ 3(−pi)^{2}+ 3(−pi) + 1= −pi

^{3}+ 3pi^{2}− 3pi + 1

**(v) 5 + 2x**

Solution:5 + 2x = 0

2x = −5

x = -5/2

Therefore remainder will be f(x) :

f(-5/2) = (-5/2)

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

= -27/8

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

^{3}− ax

^{2}+ 6x − a is divided by x – a.

Solution:Let f(x) = x

^{3}− ax^{2}+ 6x − ax − a = 0

∴ x = a

Therefore remainder will be f(x):

f(a) = (a)

^{3}− a(a^{2}) + 6(a) − a= a

^{3}− a^{3}+ 6a − a= 5a

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

^{3 }+ 7x.

Solution:7 + 3x = 0

3x = −7

x = -7/3

Therefore remainder will be f(x):

f(-7/3) = 3(-7/3)

^{3 }+ 7(-7/3)= – (343/9) + (-49/3)

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

= (-343 – 147)/9

= – 490/9 ≠ 0

∴ 7 + 3x is not a factor of 3x

^{3 }+ 7x