How do you factor $4r^2-13rt+9t^2$ ?

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Answer 1 · 2016-05-06T19:33:07

$4r^2-13rt+9t^2=(4r-9t)(r-t)$

There are several ways to see this, but since the coefficients add up to $0$ we find that $(r-t)$ is a factor, which we can show by grouping:

$4r^2-13rt+9t^2$

$=4r^2-4rt-9rt+9t^2$

$=(4r^2-4rt)-(9rt-9t^2)$

$=4r(r-t)-9t(r-t)$

$=(4r-9t)(r-t)$

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