How do you factor ${{x}^{4}}-6{{x}^{2}}-16$?

Hint: We know that the general form of a quadratic equation is $a{{x}^{2}}+bx+c$ since in this question we have a polynomial equation which cannot be factorized directly therefore, in this question we will consider the term ${{x}^{2}}$ as some variable $m$ and then solve the quadratic equation and then simplify the factors further to get the required solution.

Complete step by step solution: We have the given polynomial as:
$\Rightarrow {{x}^{4}}-6{{x}^{2}}-16$
We can rearrange the terms in the expression as:
$\Rightarrow {{\left( {{x}^{2}} \right)}^{2}}-6{{x}^{2}}-16$
Since there is no way to factorize the polynomial directly, let’s consider ${{x}^{2}}=m$. On substituting it in the polynomial, we get:
$\Rightarrow {{m}^{2}}-6m-16$
Now this expression is in the form of $a{{m}^{2}}+nx+c$therefore, we will solve this as a quadratic equation by splitting the middle term.
Therefore, the equation can be written as:
$\Rightarrow {{m}^{2}}-8m+2m-16$
on taking the common terms, we get:
$\Rightarrow m\left( m-8 \right)+2\left( m-8 \right)$
since the term $\left( m-8 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow \left( m-8 \right)\left( m+2 \right)$
Now on resubstituting the value of $m$ as ${{x}^{2}}$, we get:
$\Rightarrow \left( {{x}^{2}}-8 \right)\left( {{x}^{2}}+2 \right)$
Now we know that ${{\left( \sqrt{8} \right)}^{2}}=8$ therefore, on substituting, we get:
$\Rightarrow \left( {{x}^{2}}-{{\left( \sqrt{8} \right)}^{2}} \right)\left( {{x}^{2}}+2 \right)$
Since the term $\left( {{x}^{2}}-{{\left( \sqrt{8} \right)}^{2}} \right)$ is in the form of $\left( {{a}^{2}}-{{b}^{2}} \right)$ we can expand it further as:
$\Rightarrow \left( x-\sqrt{8} \right)\left( x+\sqrt{8} \right)\left( {{x}^{2}}+2 \right)$.
On simplifying the root terms, we get:
$\Rightarrow \left( x-2\sqrt{2} \right)\left( x+2\sqrt{2} \right)\left( {{x}^{2}}+2 \right)$
Since the term $\left( {{x}^{2}}+2 \right)$ cannot be simplified further we can say that:
$\Rightarrow {{x}^{4}}-6{{x}^{2}}-16=\left( x-2\sqrt{2} \right)\left( x+2\sqrt{2} \right)\left( {{x}^{2}}+2 \right)$, which is the required solution.

Note: In this question one root of the polynomial expression is $x=2\sqrt{2}$, which is not a rational number. This indicates that all roots of an expression are not integers, real or rational numbers therefore, the general quadratic formula should be remembered while factoring which is:
$({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the quadratic equation.