Find the remainder when \(x^3 + 3x^2 + 3x + 1\) is divided by
i)\(x + 1\)
ii)\(x - \frac{1}{2} \)
iii)\(x\)
iv)\(x + \pi\)
v)\(5 + 2x\)

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


i)The zero of \(x + 1\) is -1
[\(\because x+1= 0 \Rightarrow x = -1 \) ]

\(\therefore p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1\)
\(\Rightarrow p(-1) = -1 + 3 - 3 +1\)
\(\Rightarrow p(-1) = 0\)

Hence, By remainder theorem, required remainder = 0.


ii)The zero of \(x - \frac{1}{2} \) is x = \(\frac{1}{2} \)
[\(\because (x - \frac{1}{2} = 0 )\)]

\(\therefore p(\frac{1}{2} ) = (\frac{1}{2} )^3 + 3(\frac{1}{2} )^2 + 3(\frac{1}{2} )+ 1\)
\( \Rightarrow p(\frac{1}{2}) = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} +1\)
\(\Rightarrow p(\frac{1}{2} ) = \frac{27}{8} \)

Hence, By remainder theorem, required remainder = \( \frac{27}{8} \) .


iii)The zero of \(x\) is x = 0 [\(\because (x = 0) \) ]

\(\therefore p(0) = (0)^3 + 3(0)^2 + 3(0) + 1\)
\(\Rightarrow p(-1) = 1\)

Hence, By remainder theorem, required remainder = 1.


iv) The zero of \(x + \pi\) is \(x = -\pi\) [\(\because (x + \pi = 0)\)]

\(\therefore p(-\pi) = (-\pi)^3 + 3(-\pi)^2 + 3(-\pi) + 1\)

\(\Rightarrow p(-\pi) = -\pi^3 + 3\pi^2 - 3\pi + 1\)

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


v) The zero of \(5 + 2x\) is x = \(\frac{-5}{2} \) [\(\because (5 + 2x = 0) \)]

\(\therefore p(\frac{-5}{2} ) = (\frac{-5}{2})^3 + 3(\frac{-5}{2})^2 + 3(\frac{-5}{2} )+ 1\)
\(\Rightarrow p(\frac{-5}{2} ) = \frac{-125}{8} + \frac{75}{4} - \frac{15}{2} + 1\)
\(\Rightarrow p(\frac{-5}{2} ) = \frac{-27}{8} \)

Hence, By remainder theorem, required remainder = \(\frac{-27}{8} \).