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.

How do you factor 5m^2 - 9mn - 18n^25m^2 - 9mn - 18n^2?