1. Find the remainder when \(x^3 + 3x^2 + 3x + 1\) is divided by

i)\(x + 1\)

ii)\(x - 1/2\)

iii)\(x\)

iv)\(x + \pi\)

v)\(5 + 2x\)

i)\(x + 1\)

ii)\(x - 1/2\)

iii)\(x\)

iv)\(x + \pi\)

v)\(5 + 2x\)

Let \(p(x) = x^3 + 3x^2 + 3x + 1\)

i)The zero of \(x + 1\) is x = -1 i.e. (x + 1 = 0)

So, \(p(-1) = {-1}^3 + 3{-1}^2 + 3{-1} + 1\)

i.e. \(p(-1) = -1 + 3 - 3 +1\)

i.e.\(p(-1) = 0\)

Hence, By remainder theorem, required remainder = 0.

ii)The zero of \(x - 1/2\) is x = 1/2 i.e. (x - 1/2 = 0)

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

i.e. \(p(1/2) = 1/8 + 3/4 + 3/2 +1\)

i.e.\(p(1/2) = 27/8\)

Hence, By remainder theorem, required remainder = 27/8.

iii)The zero of \(x\) is x = 0 i.e. (x = 0)

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

i.e. \(p(-1) = 1\)

Hence, By remainder theorem, required remainder = 1.

iv)The zero of \(x + \pi\) is \(x = -\pi\) i.e. \((x + \pi = 0)\)

So, \(p(-\pi) = {-\pi}^3 + 3{-\pi}^2 + 3{-\pi} + 1\)

i.e. \(p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1\)

Hence, By remainder theorem, required remainder = \(-\pi^3 + 3\pi^2 - 3\pi + 1\).

v)The zero of \(5 + 2x\) is x = -5/2 i.e. (5 + 2x = 0)

So, \(p(-5/2) = {-5/2}^3 + 3{-5/2}^2 + 3{-5/2} + 1\)

i.e. \(p(-5/2) = -125/8 + 75/4 - 15/2 + 1\)

i.e. \(p(-5/2) = -27/8\)

Hence, By remainder theorem, required remainder = -27/8.