Question

Find the remainder when $x^3+3x^2+3x+1$ is divided by
$x-\frac{1}{2}$

Answer 1

Dividing $x^3+3x^2+3x+1$ by $x-\frac{1}{2}$

So, $x^3+3x^2+3x+1$ is dividend and $x-\frac{1}{2}$ is divisor.

Put, $x-\frac{1}{2}=0$

$\therefore$ $x=\frac{1}{2}$

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

Putting $x=\frac{1}{2}$

$p\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^3+3{\left(\frac{1}{2}\right)}^2+3\left(\frac{1}{2}\right)+1$

$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$

$=\frac{1+6+12+8}{8}$

$=\frac{27}{8}$

Thus, reminder of $p\left(\frac{1}{2}\right)=\frac{27}{8}$

$\therefore$ We will get $\frac{27}{8}$ as a reminder when we divide $x^3+3x^2+3x+1$ by $x-\frac{1}{2}$