How do you factor # x^4 - 8x^3 + 12x^2#?

2 Answers
Jul 16, 2016

Answer:

#x^2(x-6)(x-2)#

Explanation:

Take out the common factor of #x^2# first.

#x^2(x^2 -8x +12)#

Find factors of 12 which add to 8.
6 x 2 = 12 and 6 + 2 = 8. The factors we need are 6 and 2.
The signs are both negative.

#x^2(x-6)(x-2)#

Jul 16, 2016

Answer:

#x^2(x-6)(x-2)#

Explanation:

The first step in factorising is to take out the common factor #x^2#

#rArrx^2(x^2-8x+12)#

Now we require to factorise the quadratic inside the bracket.

For the standard quadratic function #color(red)(|bar(ul(color(white)(a/a)color(black)(ax^2+bx+c)color(white)(a/a)|)))#

Consider the factors which multiply to give ac and sum to give b.

For #x^2-8x+12#

a = 1 , b = -8 and c = 12

Require factors of product #ac=1xx12=12# which sum to -8

In this case these are -6 and -2 as product = 12 and sum = -8

#rArrx^2-8x+12=(x-6)(x-2)#

#rArrx^4-8x^3+12x^2=x^2(x-6)(x-2)#