Question
Find the remainder when $ {x^3} + 3{x^2} + 3x + 1$ is divided by $x - \frac{1}{2}$
✓
Answer
$x - \frac{1}{2}$
We need to find the zero of the polynomial $x - \frac{1}{2}$
$\begin{gathered} x - \frac{1}{2} = 0{\text{ }} \hfill \\ \Rightarrow {\text{ }}x = \frac{1}{2} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $x - \frac{1}{2}$ in the polynomial ${x^3} + 3{x^2} + 3x + 1$, to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} + 3\left( {\frac{1}{2}} \right) + 1$
$ = \frac{1}{8} + 3\left( {\frac{1}{4}} \right) + \frac{3}{2} + 1$
We need to find the zero of the polynomial $x - \frac{1}{2}$
$\begin{gathered} x - \frac{1}{2} = 0{\text{ }} \hfill \\ \Rightarrow {\text{ }}x = \frac{1}{2} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $x - \frac{1}{2}$ in the polynomial ${x^3} + 3{x^2} + 3x + 1$, to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} + 3\left( {\frac{1}{2}} \right) + 1$
$ = \frac{1}{8} + 3\left( {\frac{1}{4}} \right) + \frac{3}{2} + 1$
= ${1 \over 8} + {3 \over 4} + {3 \over 2} + 1$
$= \frac{{1 + 6 + 12 + 8}}{8}$
$\, = \frac{{27}}{8}$
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1 by \ x - \frac{1}{2}$ we will get the remainder as $\frac{{27}}{8}$
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