How do you factor #5m^2 - 9mn - 18n^2#?

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Answer 1 · 2015-05-25T14:32:16

Since all the terms are of order 2, factoring #5m^2-9mn-18n^2# is analogous to factoring the quadratic #5x^2-9x-18#.

This quadratic is in the form #ax^2+bx+c#, with #a=5#, #b=-9# and #c=-18#.

The discriminant is given by the formula:

#Delta = b^2-4ac = (-9)^2 - (4xx5xx-18)#

#=81+360=441=21^2#

Since this is a positive, perfect square, the quadratic #5x^2-9x-18 = 0# has two rational solutions, given by the formula:

#x = (-b+-sqrt(Delta))/(2a) = (9+-21)/10#

that is #x=3# or #x=-6/5#

These correspond to factors #(x-3)# and #(5x+6)#

These factors correspond to factors #(m-3n)# and #(5m+6n)# for the original expression.