How do you simplify square root of s^8 + square root of 25s^8 + 2 square root of s^8 - square root of s^4?

Sep 10, 2015

${s}^{2} \cdot \left(2 \sqrt{2} s + 1\right) \left(2 \sqrt{2} s - 1\right)$

Explanation:

Your starting expression looks like this

$\sqrt{{s}^{8}} + \sqrt{25 {s}^{8}} + 2 \sqrt{{s}^{8}} - \sqrt{{s}^{4}}$

You can simplify this expression by playing around a bit with some properties of radicals and exponents. For example, you could say that

$\sqrt{{s}^{8}} = \sqrt{{\left({s}^{4}\right)}^{2}} = {s}^{4}$

Likewise,

$\sqrt{{s}^{4}} = \sqrt{{\left({s}^{2}\right)}^{2}} = {s}^{2}$

The second radical term can be written as

$\sqrt{25 \cdot {s}^{8}} = \sqrt{25} \cdot \sqrt{{s}^{8}} = \sqrt{{5}^{2}} \cdot \sqrt{{\left({s}^{4}\right)}^{2}} = 5 \cdot {s}^{4}$

The expression will now take the form

${s}^{4} + 5 \cdot {s}^{4} + 2 \cdot {s}^{4} - {s}^{2}$

Group like-terms to get

$8 {s}^{4} - {s}^{2}$

You could simplify this further by using the difference of two squares factoring formula

$\textcolor{b l u e}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

In your case, you could write

$8 {s}^{4} - {s}^{2} = {s}^{2} \cdot \left(8 {s}^{2} - 1\right) = {s}^{2} \cdot \left[{\left(\sqrt{8} \cdot s\right)}^{2} - {1}^{2}\right]$

This can be simplified to

${s}^{2} \cdot \left[{\left(\sqrt{8} \cdot s\right)}^{2} - {1}^{2}\right] = {s}^{2} \cdot \left(\sqrt{8} s - 1\right) \cdot \left(\sqrt{8} s + 1\right)$

The final form of the expression will be

$\textcolor{g r e e n}{{s}^{2} \cdot \left(2 \sqrt{2} s + 1\right) \left(2 \sqrt{2} s - 1\right)}$