How do you factor #f(x)=x^3-12x^2+36x-32# completely, given that (x-2) is a factor?

1 Answer
Andrea S.
Dec 2, 2016

#x^3-12x^2+36x-32=(x-2)^2(x-8)#

Explanation:

Knowing that #(x-2)# is a factor you can divide the polynomial:

2.....|.......1.........-12.........+36.........-32
........|...................2..........-20.........+32
........|.......1.........-10..........+16...........0

and find:

#x^3-12x^2+36x-32=(x-2)(x^2-10x+16)#

We could use the quadratic formula to find the factors of #q(x)=(x^2-10x+16)#

or go simply by inspection:

1) q(x) has two changes in sign, so it has two positive real roots.
2) The product of the roots is #16#
3) the sum of the roots is #10#

It's easy to see that #x=2# and #x=8# are such roots, and thus:

#x^3-12x^2+36x-32=(x-2)^2(x-8)#

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