How do you factor $6r^3-9r^3-4r+6$ ?

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Answer 1 · 2015-06-02T20:22:21

I think the question contains an error: $-9r^3$ should be $-9r^2$

On the assumption that I'm correct, the cubic polynomial factors easily by grouping:

$6r^3-9r^2-4r+6$

$=(6r^3-9r^2)-(4r-6)$

$=3r^2(2r-3)-2(2r-3)$

$=(3r^2-2)(2r-3)$

$=(sqrt(3)r-sqrt(2))(sqrt(3)r+sqrt(2))(2r-3)$

How do you factor 6r^3-9r^3-4r+66r^3-9r^3-4r+6?