How do you factor $6 m^{2} + 43 m n - 15 n^{2}$ $6m^2+43mn-15n^2$ ?

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How do you factor $6 m^{2} + 43 m n - 15 n^{2}$ $6m^2+43mn-15n^2$ ?

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Answer 1 · 2016-06-20T07:26:04

Start by knowing that we're looking for $(A m + B n) (C m - D n)$ $(Am+Bn)(Cm-Dn)$ where $A C = 6$ $AC=6$ , $B D = - 15$ $BD=-15$ , and $A D + B C = 43$ $AD+BC=43$ , to find $(2 m + 15 n) (3 m - n)$ $(2m+15n)(3m-n)$

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:

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

We know the factors for 6 are $(1, 6), (2, 3), (- 1, - 6), and (- 2, - 3)$ $(1,6), (2,3), (-1,-6), and (-2,-3)$
We know the factors for -15 are $(1, - 15), (- 1, 15), (3, - 5), and (- 3, 5)$ $(1,-15),(-1,15),(3,-5), and (-3,5)$

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:

$(A m + B n) (C m - D n)$ $(Am+Bn)(Cm-Dn)$ where $A C = 6$ $AC=6$ , $B D = - 15$ $BD=-15$ , and $A D + B C = 43$ $AD+BC=43$

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

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

And nice - we got it on the first go.

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

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How do you factor $6 m^{2} + 43 m n - 15 n^{2}$ $6m^2+43mn-15n^2$ ?

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How do you factor $6 m^{2} + 43 m n - 15 n^{2}$ $6m^2+43mn-15n^2$ ?