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

Feb 15, 2016

${\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 4\right) \left({x}^{2} + 2 x + 4\right)$

#### 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:

$= \left({x}^{7} - 8 {x}^{4}\right) + \left(- 16 {x}^{3} + 128\right)$

Factor a common term from both sets:

$= {x}^{4} \left({x}^{3} - 8\right) - 16 \left({x}^{3} - 8\right)$

Factor an $\left({x}^{3} - 8\right)$ term from the ${x}^{4}$ and $- 16$ terms.

$= \left({x}^{3} - 8\right) \left({x}^{4} - 16\right)$

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

Differences of cubes factor as follows:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

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

${x}^{3} - 8 = \left(x + 2\right) \left({x}^{2} + x \left(2\right) + {2}^{2}\right) = \left(x + 2\right) \left({x}^{2} + 2 x + 4\right)$

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

$= \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left({x}^{4} - 16\right)$

Now, note that $\left({x}^{4} - 16\right)$ is a difference of squares. Differences of squares factor as follows:

${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

In ${x}^{4} - 16$, we see that ${x}^{4} = {\left({x}^{2}\right)}^{2}$ and $16 = {\left(4\right)}^{2}$, so $a = {x}^{2}$ and $b = 4$. This gives us a factorization of

${x}^{4} - 16 = \left({x}^{2} + 4\right) \left({x}^{2} - 4\right)$

This gives us the following factorization of the original expression:

$= \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left({x}^{2} + 4\right) \left({x}^{2} - 4\right)$

However, notice that $\left({x}^{2} - 4\right)$ is also a difference of squares. It factors into

${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right)$

Which allows us to obtain the factorization of

$= \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) \left({x}^{2} + 4\right) \left(x + 2\right) \left(x - 2\right)$

Note that there are two $\left(x - 2\right)$ terms, so this can be rewritten as

$= {\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 4\right) \left({x}^{2} + 2 x + 4\right)$