How do you factor ${{x}^{4}}-8{{x}^{2}}-9$ ?

Hint: We are given an equation ${{x}^{4}}-8{{x}^{2}}-9$ , we are asked to find all its factors. We will learn about its degree then we find a method to solve this problem. We look closely and see we have our equation in form of ${{x}^{2}}$ so we start solution by substituting ${{x}^{2}}$ as ‘t’ then we get our equation as a quadratic equation. we solve further using middle terms split. Once we split the above described quadratic to its factor then we will use $\left( \alpha +\beta \right)\left( \alpha -\beta \right)={{\alpha }^{2}}-{{\beta }^{2}}$ to factor as much as possible then at last we simplify and finish the solution.

Complete step-by-step solution:
We are given an equation ${{x}^{4}}-8{{x}^{2}}-9$. We are asked to factor the given equation.
Now we can see our equation, we are having the highest power as ‘4’ so it will have ‘4’ terms in the factor form.
As we look closely then we can see that our variable are in form of ${{x}^{2}}$ , that is we have ${{x}^{2}}$ and ${{x}^{4}}$ . as we know ${{x}^{4}}={{\left( {{x}^{2}} \right)}^{2}}$ so we can write our equation as ${{\left( {{x}^{2}} \right)}^{2}}-8{{x}^{2}}-9$ .
Now to simplify it a little we substitute ${{x}^{2}}$ as ‘t’ so we get –
${{\left( {{x}^{2}} \right)}^{2}}-8{{x}^{2}}-9={{t}^{2}}-8t-9$
Now we have a quadratic equation, we will solve this quadratic equation using the method of middle term split.
This method say equation $a{{t}^{2}}+bt+c$
We find such numbers whose product is the same as $a\times c$ and their difference or sum is the same as ‘b’.
Now in ${{t}^{2}}-8t-9$
We have $a=1,b=-8,c=-9$
So, we can see that $a\times c=1\times \left( -9 \right)=-9$
We can see that $-9\times 1=-9$
And $-9+1=-8$
So we use $-9+1$ to split middle term using these below, we get –
${{t}^{2}}-8t-9={{t}^{2}}+\left( -9+1 \right)t-9$
Opening bracket, we get –
$={{t}^{2}}-9t+t-9$
By simplifying, we get –
$=t\left( t-9 \right)+1\left( t-9 \right)$
Taking $\left( t-9 \right)$ as common, so we get –
$=\left( t+1 \right)\left( t-9 \right)$
Now we substitute ‘t’ back as ${{x}^{2}}$ , so we get –
${{x}^{4}}-8{{x}^{2}}-9=\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-9 \right)$
Now as we see ${{x}^{2}}+1$ cannot be simplified further but ${{x}^{2}}-9$ we can simplify it
We know $\left( \alpha +\beta \right)\left( \alpha -\beta \right)={{\alpha }^{2}}-{{\beta }^{2}}$
As $9={{3}^{2}}$
So we use $\alpha =x$ and $\beta =3$ so we get –
$\left( {{x}^{2}}-9 \right)=\left( {{x}^{2}}-{{3}^{2}} \right)$
Using $\left( \alpha +\beta \right)\left( \alpha -\beta \right)={{\alpha }^{2}}-{{\beta }^{2}}$ we get –
$=\left( x+3 \right)\left( x-3 \right)$
Putting it above, we get –
${{x}^{4}}-8{{x}^{2}}-9=\left( {{x}^{2}}+1 \right)\left( x+3 \right)\left( x-3 \right)$

Note: In real line we cannot factor ${{x}^{2}}+1$ , because if we do it mean we have ${{x}^{2}}+1=0$ implies $x=\sqrt{-1}$ which is not true as square roots take only positive value.
But if our equation is in a complex plane then we can solve it more further.
We know that $\sqrt{-1}=i$ so, we can split ${{x}^{2}}+1$ into linear factor as –
${{x}^{2}}+1=\left( x+i \right)\left( x-i \right)$
Hence our given equation ${{x}^{4}}-8{{x}^{2}}-9$ will have factor as $\left( x+3 \right)\left( x-3 \right)\left( x+i \right)\left( x-i \right)$ where I denotes iota $i=\sqrt{-1}$ .