# How do you factor #x^7-8x^4-16x^3+128#?

##### 1 Answer

#### Explanation:

We will employ the method of **"factor by grouping"**, in which we take a common term from the first two and last two terms of the expression.

Group the expression into two sets of two:

#=(x^7-8x^4)+(-16x^3+128)#

Factor a common term from both sets:

#=x^4(x^3-8)-16(x^3-8)#

Factor an

#=(x^3-8)(x^4-16)#

Here, we have two factoring identities. The first we'll tackle is **difference of cubes**, as both of its terms are cubed:

Differences of cubes factor as follows:

#a^3-b^3=(a-b)(a^2+ab+b^2)#

Here, we have

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

We can substitute this back into the expression we had earlier:

#=(x-2)(x^2+2x+4)(x^4-16)#

Now, note that **difference of squares**. Differences of squares factor as follows:

#a^2-b^2=(a+b)(a-b)#

In

#x^4-16=(x^2+4)(x^2-4)#

This gives us the following factorization of the original expression:

#=(x-2)(x^2+2x+4)(x^2+4)(x^2-4)#

However, notice that *also* a difference of squares. It factors into

#x^2-4=(x+2)(x-2)#

Which allows us to obtain the factorization of

#=(x-2)(x^2+2x+4)(x^2+4)(x+2)(x-2)#

Note that there are two

#=(x-2)^2(x+2)(x^2+4)(x^2+2x+4)#