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

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Answer 1 · 2016-06-20T07:26:04

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

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+43mn-15n^2#

We know the factors for 6 are #(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)#

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:

#(Am+Bn)(Cm-Dn)# where #AC=6#, #BD=-15#, and #AD+BC=43#

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

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

And nice - we got it on the first go.

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