# How do you factor x^3-8x^2+19x-12?

Apr 9, 2015
• We have the polynomial color(red)(p(x) = x^3-8x^2+19x-12
As the degree of the polynomial is greater than 2, we need to use Synthetic Division to factorize it.

• In Synthetic Division, we first find a value of $x$ that makes this polynomial equal to zero. That is, we need to find the ZERO of the polynomial. We do that by Trial and Error.
For $x = \textcolor{b l u e}{1}$, this Polynomial will equal zero.

$p \left(1\right) = {1}^{3} - 8 \cdot {1}^{2} + 19 \cdot 1 - 12 = 1 - 8 + 19 - 12 = 0$

As $x = 1$ makes this polynomial zero, we know for sure that $\textcolor{g r e e n}{\left(x - 1\right)}$ is one of the factors of the polynomial.

• To find the other two factors, we use the process of Synthetic Division. To explain the entire process of Synthetic Division using text is a bit difficult, but this image should help:

The 1, -7 and 12 give us another factor of the polynomial:
${x}^{2} - 7 x + 12$

We have reduced p(x) to

$p \left(x\right) = \left(x - 1\right) \left({x}^{2} - 7 x + 12\right)$

But ${x}^{2} - 7 x + 12$ can be factorised further
$= {x}^{2} - 3 x - 4 x + 12$
$= x \left(x - 3\right) - 4 \left(x - 3\right)$
$= \left(x - 3\right) \left(x - 4\right)$

So finally, we get

$p \left(x\right) = \left(x - 1\right) \left(x - 3\right) \left(x - 4\right)$

x^3-8x^2+19x-12 = color(green)( (x-1)(x-3)(x-4)