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

1 Answer
mason m
Feb 15, 2016

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

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)# term from the #x^4# and #-16# terms.

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

Here, we have two factoring identities. The first we'll tackle is #(x^3-8)#, which is a difference of cubes, as both of its terms are cubed: #x^3=(x)^3# and #8=(2)^3#.

Differences of cubes factor as follows:

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

Here, we have #a=x# and #b=2#, so

#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 #(x^4-16)# is a difference of squares. Differences of squares factor as follows:

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

In #x^4-16#, we see that #x^4=(x^2)^2# and #16=(4)^2#, so #a=x^2# and #b=4#. This gives us a factorization of

#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 #(x^2-4)# is 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)# terms, so this can be rewritten as

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

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