# How do you factor completely x^2 - 6x + 8 ?

Apr 29, 2016

For ${x}^{2} - 6 x + 8 = 0$
$0 = {x}^{2} - 4 x - 2 x + 8$
$0 = x \left(x - 4\right) - 2 \left(x - 4\right)$
$0 = \left(x - 4\right) \left(x - 2\right)$

$\left(x - 4\right) = 0$ OR $\left(x - 2\right) = 0$
$x = 4$ OR $x = 2$

For $x = 4 \mathmr{and} 2$
${x}^{2} - 6 x + 8$ is zero
Hence $4 \mathmr{and} 2$ are its factors