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

##### 1 Answer

#### Answer:

#### Explanation:

First note that all of the terms are divisible by

#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

Note also that the pattern of the signs of the coefficients of this cubic is

Further note that

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

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

#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

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)#