Question

What is the completely factored form of f(x)=x3+5x2+4x−6?

Answer 1

$f(x)=x^3+5x^2+4x-6$
$f(-3)=(-3)^3+5(-3)^2+4(-3)-6=0\implies x+3\text{ is a factor of }f(x)$

Synthetic division yields

-3  |  1   5   4   -6
.    |      -3  -6    6
- - - - - - - - - - - - -
.    |  1   2  -2    0

which translates to

$\dfrac{x^3+5x^2+4x-6}{x+3}=x^2+2x-2$

with remainder 0. Now by the quadratic formula,

$x^2+2x-2=0\implies x=\dfrac{-2\pm\sqrt{2^2-4(1)(-2)}}2=-1\pm\sqrt3$

and so

$f(x)=x^3+5x^2+4x-6=(x+3)(x-(-1+\sqrt3))(x-(-1-\sqrt3))$