# How do you factor completely 4x^3-16x^2-9x+36?

Feb 1, 2017

$4 {x}^{3} - 16 {x}^{2} - 9 x + 36 = \left(2 x - 3\right) \left(2 x + 3\right) \left(x - 4\right)$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We will use this with $a = 2 x$ and $b = 3$, but first note that the ratio of the first and second terms of the given cubic is the same as that between the third and fourth terms. So this cubic factors by grouping:

$4 {x}^{3} - 16 {x}^{2} - 9 x + 36 = \left(4 {x}^{3} - 16 {x}^{2}\right) - \left(9 x - 36\right)$

$\textcolor{w h i t e}{4 {x}^{3} - 16 {x}^{2} - 9 x + 36} = 4 {x}^{2} \left(x - 4\right) - 9 \left(x - 4\right)$

$\textcolor{w h i t e}{4 {x}^{3} - 16 {x}^{2} - 9 x + 36} = \left(4 {x}^{2} - 9\right) \left(x - 4\right)$

$\textcolor{w h i t e}{4 {x}^{3} - 16 {x}^{2} - 9 x + 36} = \left({\left(2 x\right)}^{2} - {3}^{2}\right) \left(x - 4\right)$

$\textcolor{w h i t e}{4 {x}^{3} - 16 {x}^{2} - 9 x + 36} = \left(2 x - 3\right) \left(2 x + 3\right) \left(x - 4\right)$