Question

Find the remainder when $x^3+3x^2+3x+1$ is divided by (i) $(x+1)$ and (ii) $x-\frac{1}{2}$ [ 5 marks]

Answer 1

We know that, if a polynomial $p\left(x\right)$ is divided by $x-a$ then the remainder is $p\left(a\right)$. ( 1 Mark )

Now, for $(x+1)$

$p\left(x\right)=p(-1)=(-1{)}^3+3\times (-1{)}^2+3\times (-1)+1$

$p(-1)=-1+3-3+1$

$p(-1)=0$

Hence by the remainder theorem, 0 is the remainder when $x^3+3x^2+3x+11$ is divided by $(x+1)$ ( 2 Mark )

Similarly for $x-\frac{1}{2}$

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

$p\left(\frac{1}{2}\right)=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$

$p\left(\frac{1}{2}\right)=\frac{(1+6+12+8)}{8}$

$p\left(\frac{1}{2}\right)=\frac{27}{8}$

Hence by the remainder theorem, $\frac{27}{8}$ is the remainder when $x^3+3x^2+3x+1$ is divided by $(x-\frac{1}{2})$ ( 2 Mark )