# How do you factor completely 5y^8 - 125?

Apr 24, 2016

$5 {y}^{8} - 125$

$= 5 \left({y}^{4} - 5\right) \left({y}^{4} + 5\right)$

$= 5 \left(y - \sqrt{5}\right) \left(y + \sqrt{5}\right) \left({y}^{2} + \sqrt{5}\right) \left({y}^{2} - \sqrt{20} y + \sqrt{5}\right) \left({y}^{2} + \sqrt{20} y + \sqrt{5}\right)$

#### Explanation:

First note that both terms are divisible by $5$, so separate that out as a factor:

$5 {y}^{8} - 125 = 5 \left({y}^{8} - 25\right)$

Next we will use the difference of squares identity, which may be written:

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

with $a = {y}^{4}$ and $b = 5$, as follows:

${y}^{8} - 25 = {\left({y}^{4}\right)}^{2} - {5}^{2} = \left({y}^{4} - 5\right) \left({y}^{4} + 5\right)$

Next use the difference of squares identity again, this time with $a = {y}^{2}$ and $b = \sqrt{5}$ as follows:

${y}^{4} - 5 = {\left({y}^{2}\right)}^{2} - {\left(\sqrt{5}\right)}^{2} = \left({y}^{2} - \sqrt{5}\right) \left({y}^{2} + \sqrt{5}\right)$

Next use the difference of squares identity with $a = y$ and $b = \sqrt{5}$ as follows:

${y}^{2} - \sqrt{5} = {y}^{2} - {\left(\sqrt{5}\right)}^{2} = \left(y - \sqrt{5}\right) \left(y + \sqrt{5}\right)$

Next we will factor $\left({y}^{4} + 5\right)$

First notice that:

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

In particular, if $k = \sqrt{2}$ then

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

Let $a = y$, $b = \sqrt{5}$ and $k = \sqrt{2}$.

Then:

${y}^{4} + 5 = {y}^{4} + {\left(\sqrt{5}\right)}^{4}$

$= \left({y}^{2} - \sqrt{2} \sqrt{5} y + \sqrt{5}\right) \left({y}^{2} + \sqrt{2} \sqrt{5} y + \sqrt{5}\right)$

$= \left({y}^{2} - \sqrt{20} y + \sqrt{5}\right) \left({y}^{2} + \sqrt{20} y + \sqrt{5}\right)$

Putting it all together:

$5 {y}^{8} - 125 = 5 \left(y - \sqrt{5}\right) \left(y + \sqrt{5}\right) \left({y}^{2} + \sqrt{5}\right) \left({y}^{2} - \sqrt{20} y + \sqrt{5}\right) \left({y}^{2} + \sqrt{20} y + \sqrt{5}\right)$