# Factor 36x^3+12x^2-72x-24 ?

Dec 21, 2016

$36 {x}^{3} + 12 {x}^{2} - 72 x - 24 = 12 \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left(3 x + 1\right)$

#### Explanation:

Note that the ratio of the $1$st and $2$nd terms is the same as that between the $3$rd and $4$th terms. So this cubic will factor by grouping. Also note that all of the coefficients are divisible by $12$, so we can separate that out as a factor first:

$36 {x}^{3} + 12 {x}^{2} - 72 x - 24 = 12 \left(3 {x}^{3} + {x}^{2} - 6 x - 2\right)$

$\textcolor{w h i t e}{36 {x}^{3} + 12 {x}^{2} - 72 x - 24} = 12 \left(\left(3 {x}^{3} + {x}^{2}\right) - \left(6 x + 2\right)\right)$

$\textcolor{w h i t e}{36 {x}^{3} + 12 {x}^{2} - 72 x - 24} = 12 \left({x}^{2} \left(3 x + 1\right) - 2 \left(3 x + 1\right)\right)$

$\textcolor{w h i t e}{36 {x}^{3} + 12 {x}^{2} - 72 x - 24} = 12 \left({x}^{2} - 2\right) \left(3 x + 1\right)$

$\textcolor{w h i t e}{36 {x}^{3} + 12 {x}^{2} - 72 x - 24} = 12 \left({x}^{2} - {\left(\sqrt{2}\right)}^{2}\right) \left(3 x + 1\right)$

$\textcolor{w h i t e}{36 {x}^{3} + 12 {x}^{2} - 72 x - 24} = 12 \left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) \left(3 x + 1\right)$