Question

Find the factors of the polynomial $(x^3+12x^2+48x+64)$
using synthetic division.

Answer 1

$p\left(x\right)=(x^3+12x^2+48x+64)$
Using trial and error method,
when
$x=-4p(-4)=(-4{)}^3+12\times (-4{)}^2+48\times (-4)+64)=-64+192-192+64=0$
Hence, one factor = $(x+4)$

Using synthetic division,
$\begin{array}{ccccc}& x^3& x^2& x^1& x^0\\ & 1& 12& 48& 64\\ -4& \downarrow & -4& -32& -64\\ & 1& 18& 16& |-0\end{array}$

$\therefore p\left(x\right)=(x+4)(x^2+8x+16)$

Now, using trial and error method,
when,
$x=-4q\left(x\right)=(x^2+8x+16)=(-4{)}^2+8\times (-4)+16=0$
Hence, another factor is $(x+4)$

$\begin{array}{cccc}& x^2& x^1& x^0\\ & 1& 8& 16\\ -4& \downarrow & -4& -16\\ & 1& 4& |-0\end{array}$

$\therefore p\left(x\right)=(x+4)(x+4)(x+4)=(x+4{)}^3$