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)#