How do you factor #15x^3+7x^2-40x+12#?

1 Answer
George C.
May 9, 2016

#15x^3+7x^2-40x+12 = (3x-1)(5x-6)(x+2)#

Explanation:

#f(x) = 15x^3+7x^2-40x+12#

By the rational root theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #12# and #q# a divisor of the coefficient #15# of the leading term.

That means that the only possible rational zeros are:

#+-1/15#, #+-2/15#, #+-1/5#, #+-4/15#, #+-1/3#, #+-2/5#, #+-2/3#, #+-3/5#, #+-4/5#, #+-1#, #+-6/5#, #+-4/3#, #+-2#, #+-12/5#, #+-3#, #+-4#, #+-6#, #+-12#

Trying each in turn, the first one that works is #x=1/3#:

#f(1/3) = 15/27+7/9-40/3+12 = (5+7-120+108)/9 = 0#

So #(3x-1)# is a factor:

#15x^3+7x^2-40x+12 = (3x-1)(5x^2+4x-12)#

To factor the remaining quadratic, use an AC method.

Find a pair of factors of #AC = 5*12 = 60# which differ by #B=4#

The pair #10, 6# works.

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

#5x^2+4x-12#

#=5x^2+10x-6x-12#

#=(5x^2+10x)-(6x+12)#

#=5x(x+2)-6(x+2)#

#=(5x-6)(x+2)#

Putting it all together:

#15x^3+7x^2-40x+12 = (3x-1)(5x-6)(x+2)#

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