Question

Find the reminder when $x^3+3x^3+3x+1$ is divided by
a) $5+2x$
b) $x-\frac{1}{2}$

Answer 1

Considering the question to be,

Find the reminder when $x^3+3x^3+3x+1$ is divided by

(i)$5+2x$

(ii)$x-\frac{1}{2}$

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

(i)$5+2x$

Put divisor =0

$5+2x=0$

$\Rightarrow x=-\frac{5}{2}$

Putting $x=-\frac{5}{2}$ in $x^3+3x^3+3x+1$

we get,

$p(-\frac{5}{2})={(-\frac{5}{2})}^3+3{(-\frac{5}{2})}^3+3(-\frac{5}{2})+1$

$=-\left(\frac{125}{8}\right)-\left(\frac{375}{8}\right)-\left(\frac{15}{2}\right)+1$

$=\frac{-125-375-60+8}{8}$

$=\frac{-552}{8}=-69$

Thus, remainder of

$p(-\frac{5}{2})=-69$

(ii)$x-\frac{1}{2}$

Put divisor =0

$x\dfrac{1}{2}=0$

$\Rightarrow x=-\frac{1}{2}$

Putting $x=\frac{1}{2}$ in $x^3+3x^3+3x+1$

we get,

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

$=\left(\frac{1}{8}\right)+\left(\frac{3}{8}\right)+\left(\frac{3}{2}\right)+1$

$=\frac{1+3+12+8}{8}$

$=\frac{24}{8}=3$

Thus, remainder of

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