# How do you factor 8x^3+4x^2-18x-9?

Jun 22, 2016

$8 {x}^{3} + 4 {x}^{2} - 18 x - 9 = \left(2 x - 3\right) \left(2 x + 3\right) \left(2 x + 1\right)$

#### Explanation:

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms. So we can factor this cubic by grouping:

$8 {x}^{3} + 4 {x}^{2} - 18 x - 9$

$= \left(8 {x}^{3} + 4 {x}^{2}\right) - \left(18 x + 9\right)$

$= 4 {x}^{2} \left(2 x + 1\right) - 9 \left(2 x + 1\right)$

$= \left(4 {x}^{2} - 9\right) \left(2 x + 1\right)$

$= \left({\left(2 x\right)}^{2} - {3}^{2}\right) \left(2 x + 1\right)$

$= \left(2 x - 3\right) \left(2 x + 3\right) \left(2 x + 1\right)$

Note that we also used the difference of squares identity:

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

with $a = 2 x$ and $b = 3$