# How do you factor 16-a^16?

Apr 3, 2018

(4-${a}^{8}$)(4+a^8)

#### Explanation:

Difference of two squares

Apr 3, 2018

$16 - {a}^{16} = \left(\sqrt[4]{2} - a\right) \left(\sqrt[4]{2} + a\right) \left(\sqrt{2} + {a}^{2}\right) \left(\sqrt{2} - {2}^{\frac{3}{4}} a + {a}^{2}\right) \left(\sqrt{2} + {2}^{\frac{3}{4}} a + {a}^{2}\right) \left(\sqrt{2} - \sqrt{2 \sqrt{2} + 2} a + {a}^{2}\right) \left(\sqrt{2} + \sqrt{2 \sqrt{2} + 2} a + {a}^{2}\right) \left(\sqrt{2} - \sqrt{2 \sqrt{2} - 2} a + {a}^{2}\right) \left(\sqrt{2} + \sqrt{2 \sqrt{2} - 2} a + {a}^{2}\right)$

#### Explanation:

The difference of squares identity can be written:

${A}^{2} - {B}^{2} = \left(A - B\right) \left(A + B\right)$

Given:

$16 - {a}^{16}$

We can start to factor this using the difference of squares identity as follows:

$16 - {a}^{16} = {4}^{2} - {\left({a}^{8}\right)}^{2}$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left(4 - {a}^{8}\right) \left(4 + {a}^{8}\right)$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left({2}^{2} - {\left({a}^{4}\right)}^{2}\right) \left(4 + {a}^{8}\right)$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left(2 - {a}^{4}\right) \left(2 + {a}^{4}\right) \left(4 + {a}^{8}\right)$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left({\left(\sqrt{2}\right)}^{2} - {\left({a}^{2}\right)}^{2}\right) \left(2 + {a}^{4}\right) \left(4 + {a}^{8}\right)$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left(\sqrt{2} - {a}^{2}\right) \left(\sqrt{2} + {a}^{2}\right) \left(2 + {a}^{4}\right) \left(4 + {a}^{8}\right)$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left({\left(\sqrt[4]{2}\right)}^{2} - {a}^{2}\right) \left(\sqrt{2} + {a}^{2}\right) \left(2 + {a}^{4}\right) \left(4 + {a}^{8}\right)$

$\textcolor{w h i t e}{16 - {a}^{16}} = \left(\sqrt[4]{2} - a\right) \left(\sqrt[4]{2} + a\right) \left(\sqrt{2} + {a}^{2}\right) \left(2 + {a}^{4}\right) \left(4 + {a}^{8}\right)$

Sticking with real coefficients for now, we can go a little further with some of the remaining factors.

For example:

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

$\textcolor{w h i t e}{2 + {a}^{4}} = {\left(\sqrt{2} + {a}^{2}\right)}^{2} - {\left({2}^{\frac{3}{4}} a\right)}^{2}$

$\textcolor{w h i t e}{2 + {a}^{4}} = \left(\left(\sqrt{2} + {a}^{2}\right) - {2}^{\frac{3}{4}} a\right) \left(\left(\sqrt{2} + {a}^{2}\right) + {2}^{\frac{3}{4}} a\right)$

$\textcolor{w h i t e}{2 + {a}^{4}} = \left(\sqrt{2} - {2}^{\frac{3}{4}} a + {a}^{2}\right) \left(\sqrt{2} + {2}^{\frac{3}{4}} a + {a}^{2}\right)$

Similarly:

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

$\textcolor{w h i t e}{4 + {a}^{8}} = {\left(2 + {a}^{4}\right)}^{2} - {\left(2 {a}^{2}\right)}^{2}$

$\textcolor{w h i t e}{4 + {a}^{8}} = \left(\left(2 + {a}^{4}\right) - 2 {a}^{2}\right) \left(\left(2 + {a}^{4}\right) + 2 {a}^{2}\right)$

$\textcolor{w h i t e}{4 + {a}^{8}} = \left(2 - 2 {a}^{2} + {a}^{4}\right) \left(2 + 2 {a}^{2} + {a}^{4}\right)$

Then:

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

$\textcolor{w h i t e}{2 - 2 {a}^{2} + {a}^{4}} = {\left(\sqrt{2} + {a}^{2}\right)}^{2} - {\left(\sqrt{2 \sqrt{2} + 2} a\right)}^{2}$

$\textcolor{w h i t e}{2 - 2 {a}^{2} + {a}^{4}} = \left(\left(\sqrt{2} + {a}^{2}\right) - \sqrt{2 \sqrt{2} + 2} a\right) \left(\left(\sqrt{2} + {a}^{2}\right) + \sqrt{2 \sqrt{2} + 2} a\right)$

$\textcolor{w h i t e}{2 - 2 {a}^{2} + {a}^{4}} = \left(\sqrt{2} - \sqrt{2 \sqrt{2} + 2} a + {a}^{2}\right) \left(\sqrt{2} + \sqrt{2 \sqrt{2} + 2} a + {a}^{2}\right)$

Similarly:

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

We can combine all of these to give a complete factorisation of $16 - {a}^{16}$ over the reals:

$16 - {a}^{16} = \left(\sqrt[4]{2} - a\right) \left(\sqrt[4]{2} + a\right) \left(\sqrt{2} + {a}^{2}\right) \left(\sqrt{2} - {2}^{\frac{3}{4}} a + {a}^{2}\right) \left(\sqrt{2} + {2}^{\frac{3}{4}} a + {a}^{2}\right) \left(\sqrt{2} - \sqrt{2 \sqrt{2} + 2} a + {a}^{2}\right) \left(\sqrt{2} + \sqrt{2 \sqrt{2} + 2} a + {a}^{2}\right) \left(\sqrt{2} - \sqrt{2 \sqrt{2} - 2} a + {a}^{2}\right) \left(\sqrt{2} + \sqrt{2 \sqrt{2} - 2} a + {a}^{2}\right)$