# How do you factor completely 12y^2+9y+8y+6?

Nov 19, 2015

Factor by grouping to find:

$12 {y}^{2} + 9 y + 8 y + 6 = \left(3 y + 2\right) \left(4 y + 3\right)$

#### Explanation:

$12 {y}^{2} + 9 y + 8 y + 6$

$= \left(12 {y}^{2} + 9 y\right) + \left(8 y + 6\right)$

$= 3 y \left(4 y + 3\right) + 2 \left(4 y + 3\right)$

$= \left(3 y + 2\right) \left(4 y + 3\right)$

Note that the hard work / fun has already been done for you in the split $9 y + 8 y$. Normally you would be given a quadratic like $12 {y}^{2} + 17 y + 6$ to factor and have to find the split yourself.

If you are given something like $12 {y}^{2} + 17 y + 6$ to factor, then one way is to multiply $12 \times 6 = 72$ then try to find a pair of factors that multiply to give $72$ and sum to give $17$. Once you find the pair $9$, $8$ works then you can split $17 y = 9 y + 8 y$ and factor by grouping as we have done here.