# How do you factor 6m^2+43mn-15n^2 ?

Start by knowing that we're looking for $\left(A m + B n\right) \left(C m - D n\right)$ where $A C = 6$, $B D = - 15$, and $A D + B C = 43$, to find $\left(2 m + 15 n\right) \left(3 m - n\right)$

#### Explanation:

The key to this kind of problem is to find factors for 6 and -15 that when multiplied and added all together, adds to 43. So let's do that.

Starting with the original:

${6}^{2} + 43 m n - 15 {n}^{2}$

We know the factors for 6 are $\left(1 , 6\right) , \left(2 , 3\right) , \left(- 1 , - 6\right) , \mathmr{and} \left(- 2 , - 3\right)$
We know the factors for -15 are $\left(1 , - 15\right) , \left(- 1 , 15\right) , \left(3 , - 5\right) , \mathmr{and} \left(- 3 , 5\right)$

We also know that there will be an m in the first term and an n in the second.

Overall, we're looking to find a solution in the form of:

$\left(A m + B n\right) \left(C m - D n\right)$ where $A C = 6$, $B D = - 15$, and $A D + B C = 43$

We need a pretty big positive number for $A D + B C = 43$, so I'm going to first look at using B=15:

$A = 2 , B = 15 , C = 3 , D = - 1$, so $A C = 6 , B D = - 15 , A D + B C = - 2 + 45 = 43$

And nice - we got it on the first go.

So we have $\left(2 m + 15 n\right) \left(3 m - n\right)$