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Question: How do you factor \({{x}^{4}}-8{{x}^{2}}-9\) ?...

How do you factor x4−8x2−9{{x}^{4}}-8{{x}^{2}}-9 ?

Explanation

Solution

We are given an equation x4−8x2−9{{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 x2{{x}^{2}} so we start solution by substituting x2{{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 (α+β)(α−β)=α2−β2\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 x4−8x2−9{{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 x2{{x}^{2}} , that is we have x2{{x}^{2}} and x4{{x}^{4}} . as we know x4=(x2)2{{x}^{4}}={{\left( {{x}^{2}} \right)}^{2}} so we can write our equation as (x2)2−8x2−9{{\left( {{x}^{2}} \right)}^{2}}-8{{x}^{2}}-9 .
Now to simplify it a little we substitute x2{{x}^{2}} as ‘t’ so we get –
(x2)2−8x2−9=t2−8t−9{{\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 at2+bt+ca{{t}^{2}}+bt+c
We find such numbers whose product is the same as a×ca\times c and their difference or sum is the same as ‘b’.
Now in t2−8t−9{{t}^{2}}-8t-9
We have a=1,b=−8,c=−9a=1,b=-8,c=-9
So, we can see that a×c=1×(−9)=−9a\times c=1\times \left( -9 \right)=-9
We can see that −9×1=−9-9\times 1=-9
And −9+1=−8-9+1=-8
So we use −9+1-9+1 to split middle term using these below, we get –
t2−8t−9=t2+(−9+1)t−9{{t}^{2}}-8t-9={{t}^{2}}+\left( -9+1 \right)t-9
Opening bracket, we get –
=t2−9t+t−9={{t}^{2}}-9t+t-9
By simplifying, we get –
=t(t−9)+1(t−9)=t\left( t-9 \right)+1\left( t-9 \right)
Taking (t−9)\left( t-9 \right) as common, so we get –
=(t+1)(t−9)=\left( t+1 \right)\left( t-9 \right)
Now we substitute ‘t’ back as x2{{x}^{2}} , so we get –
x4−8x2−9=(x2+1)(x2−9){{x}^{4}}-8{{x}^{2}}-9=\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-9 \right)
Now as we see x2+1{{x}^{2}}+1 cannot be simplified further but x2−9{{x}^{2}}-9 we can simplify it
We know (α+β)(α−β)=α2−β2\left( \alpha +\beta \right)\left( \alpha -\beta \right)={{\alpha }^{2}}-{{\beta }^{2}}
As 9=329={{3}^{2}}
So we use α=x\alpha =x and β=3\beta =3 so we get –
(x2−9)=(x2−32)\left( {{x}^{2}}-9 \right)=\left( {{x}^{2}}-{{3}^{2}} \right)
Using (α+β)(α−β)=α2−β2\left( \alpha +\beta \right)\left( \alpha -\beta \right)={{\alpha }^{2}}-{{\beta }^{2}} we get –
=(x+3)(x−3)=\left( x+3 \right)\left( x-3 \right)
**Putting it above, we get –
x4−8x2−9=(x2+1)(x+3)(x−3){{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 x2+1{{x}^{2}}+1 , because if we do it mean we have x2+1=0{{x}^{2}}+1=0 implies x=−1x=\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 −1=i\sqrt{-1}=i so, we can split x2+1{{x}^{2}}+1 into linear factor as –
x2+1=(x+i)(x−i){{x}^{2}}+1=\left( x+i \right)\left( x-i \right)
Hence our given equation x4−8x2−9{{x}^{4}}-8{{x}^{2}}-9 will have factor as (x+3)(x−3)(x+i)(x−i)\left( x+3 \right)\left( x-3 \right)\left( x+i \right)\left( x-i \right) where I denotes iota i=−1i=\sqrt{-1} .