# How do you factor f(x)=x^3-12x^2+36x-32 completely, given that (x-2) is a factor?

Dec 2, 2016

${x}^{3} - 12 {x}^{2} + 36 x - 32 = {\left(x - 2\right)}^{2} \left(x - 8\right)$

#### Explanation:

Knowing that $\left(x - 2\right)$ is a factor you can divide the polynomial:

2.....|.......1.........-12.........+36.........-32
........|...................2..........-20.........+32
........|.......1.........-10..........+16...........0

and find:

${x}^{3} - 12 {x}^{2} + 36 x - 32 = \left(x - 2\right) \left({x}^{2} - 10 x + 16\right)$

We could use the quadratic formula to find the factors of $q \left(x\right) = \left({x}^{2} - 10 x + 16\right)$

or go simply by inspection:

1) q(x) has two changes in sign, so it has two positive real roots.
2) The product of the roots is $16$
3) the sum of the roots is $10$

It's easy to see that $x = 2$ and $x = 8$ are such roots, and thus:

${x}^{3} - 12 {x}^{2} + 36 x - 32 = {\left(x - 2\right)}^{2} \left(x - 8\right)$