# How do you factor 6y^2- 58xy - 168 x^2?

May 16, 2015

$6 {y}^{2} - 58 x y - 168 {x}^{2} = 2 \left(3 y + 7 x\right) \left(y - 12 x\right)$

To find this, first notice the common factor 2 and factor it out

$6 {y}^{2} - 58 x y - 168 {x}^{2} = 2 \left(3 {y}^{2} - 29 x y - 84 {x}^{2}\right)$

If $3 {y}^{2} - 29 x y - 84 {x}^{2}$ has linear factors with integer coefficients then they must take the form $\left(3 y + a x\right) \left(y + b x\right)$ in order that the coefficient of ${y}^{2}$ is $3$.

$\left(3 y + a x\right) \left(y + b x\right) = 3 {y}^{2} + \left(a + 3 b\right) x y + a b {x}^{2}$

Comparing coefficients of $x y$ and ${x}^{2}$, we have:

$a + 3 b = - 29$ and $a b = - 84$

From $a + 3 b = - 29$ we can tell that one of $a$ and $b$ is odd and the other even.

$a b = - 84$ has factors $7$ and $12$, which are odd and even. So we can try $\pm 7$ and $\pm 12$ as values of $a$ and $b$ in some order.

We can quickly find that if $a = 7$ and $b = - 12$ then $a + 3 b = - 29$.

Hence $3 {y}^{2} - 29 x y - 84 {x}^{2} = \left(3 y + 7 x\right) \left(y - 12 x\right)$

and $6 {y}^{2} - 58 x y - 168 {x}^{2} = 2 \left(3 y + 7 x\right) \left(y - 12 x\right)$.