# How do you factor completely 24y^3+56y^2-6y-14 ?

Mar 10, 2016

2(3y+7)(2y-1)(2y+1)

#### Explanation:

Begin by 'grouping' the expression

$\left[24 {y}^{3} + 56 {y}^{2}\right] + \left[- 6 y - 14\right]$

now factor each group

$\Rightarrow 8 {y}^{2} \left(3 y + 7\right) - 2 \left(3 y + 7\right)$

there is a common factor of (3y + 7 )

$\Rightarrow \left(3 y + 7\right) \left(8 {y}^{2} - 2\right)$

common factor of 2 in $\left(8 {y}^{2} - 2\right) = 2 \left(4 {y}^{2} - 1\right)$

$4 {y}^{2} - 1 \text{ is a difference of squares }$

and $4 {y}^{2} - 1 = \left(2 y - 1\right) \left(2 y + 1\right)$

Finally it all comes together as $2 \left(3 y + 7\right) \left(2 y - 1\right) \left(2 y + 1\right)$