How do you factor #15x^4-52x^3+52x^2-16x#?

1 Answer
Jul 20, 2017

Answer:

#15x^4-52x^3+52x^2-16x = x(x-2)(3x-2)(5x-4)#

Explanation:

First note that all of the terms are divisible by #x#, so we can separate that out as a factor:

#15x^4-52x^3+52x^2-16x = x(15x^3-52x^2+52x-16)#

By the rational roots theorem, any rational zeros of the remaining cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #16# and #q# a divisor of the coefficient #15# of the leading term.

Note also that the pattern of the signs of the coefficients of this cubic is #+ - + -#. By Descartes' Rule of Signs, since this changes sign #3# times, the cubic has #3# or #1# positive real zeros. Reversing the signs on the terms of odd degree we get the pattern #- - - -#. With no changes of sign we can deduce that the cubic has no negative real zeros.

Further note that #15=3*5# has only two factors greater than #1#. So if all of the three zeros of the cubic are rational, then at least one of them is an integer.

So first look for positive, integer zeros, which must be factors of #16#.

We find:

#15(color(blue)(1))^3-52(color(blue)(1))^2+52(color(blue)(1))-16 = 15-52+52-16 = -1#

#15(color(blue)(2))^3-52(color(blue)(2))^2+52(color(blue)(2))-16 = 120-208+104-16 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#15x^3-52x^2+52x-16 = (x-2)(15x^2-22x+8)#

To factor the remaining quadratic use an AC method:

Find a pair of factors of #AC=15*8 = 120# with sum #B=22#. The pair #12, 10# works.

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

#15x^2-22x+8 = (15x^2-12x)-(10x-8)#

#color(white)(15x^2-22x+8) = 3x(5x-4)-2(5x-4)#

#color(white)(15x^2-22x+8) = (3x-2)(5x-4)#

Putting it all together:

#15x^4-52x^3+52x^2-16x = x(x-2)(3x-2)(5x-4)#