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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)\(5 + 2x\)
Answer :

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.