Is it possible to factor #f(y)=2y^3+7y^2-30y #? If so, what are the factors?

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Answer 1 · 2017-11-08T21:24:20

#f(y) = y(2y-5)(y+6)#

Given:

#f(y) = 2y^3+7y^2-30y#

Note that all of the terms are divisible by #y#. So we can separate that out as a factor:

#2y^3+7y^2-30y = y(2y^2+7y-30)#

Next try an AC method to factor the remaining quadratic:

Find a pair of factors of #AC = 2*30 = 60# which differ by #B=7#

(We look for difference rather than sum since the coefficient of the constant term is negative)

The pair #12, 5# works.

Use this pair to split the middle term and factor by grouping:

#2y^2+7y-30 = (2y^2+12y)-(5y+30)#

#color(white)(2y^2+7y-30) = 2y(y+6)-5(y+6)#

#color(white)(2y^2+7y-30) = (2y-5)(y+6)#

So:

#f(y) = y(2y-5)(y+6)#