Consider $x^{6}-25x^{4}-16x^{2}+400$. Do the grouping $x^{6}-25x^{4}-16x^{2}+400=\left(x^{6}-25x^{4}\right)+\left(-16x^{2}+400\right)$, and factor out $x^{4}$ in the first and $-16$ in the second group.
Factor out common term $x^{2}-25$ by using distributive property.
$$\left(x^{2}-25\right)\left(x^{4}-16\right)$$
Consider $x^{2}-25$. Rewrite $x^{2}-25$ as $x^{2}-5^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x-5\right)\left(x+5\right)$$
Consider $x^{4}-16$. Rewrite $x^{4}-16$ as $\left(x^{2}\right)^{2}-4^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x^{2}-4\right)\left(x^{2}+4\right)$$
Consider $x^{2}-4$. Rewrite $x^{2}-4$ as $x^{2}-2^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
$$\left(x-2\right)\left(x+2\right)$$
Rewrite the complete factored expression. Polynomial $x^{2}+4$ is not factored since it does not have any rational roots.
