Question

How do you factor \[{{x}^{4}}-{{x}^{2}}-20?\]

Answer 1

Hint: We are given an equation \[{{x}^{4}}-{{x}^{2}}-20\] and we are asked to find all its factors. We will learn about its degree and then we will find the method to solve this problem. We will look closely and see we have our equation in the form of \[{{x}^{2}}\] and so we start the solution by substituting \[{{x}^{2}}\] as t and then we get our equation as a quadratic equation. We will solve further using the middle terms split, once 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 will simplify and finalize the solution.

Complete step-by-step solution:
We are given an equation \[{{x}^{2}}-x-20.\] We are asked to factor the given equation. Now we can see that in 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 variables are in the form of \[{{x}^{2}},\] that is we have \[{{x}^{2}}\] and \[{{x}^{4}}.\] As we know that \[{{x}^{4}}={{\left( {{x}^{2}} \right)}^{2}},\] so we can write our equation as \[{{\left( {{x}^{2}} \right)}^{2}}-{{x}^{2}}-20.\] Now, to simplify it a little we will substitute \[{{x}^{2}}\] as t. So, we get,
\[{{\left( {{x}^{2}} \right)}^{2}}-{{x}^{2}}-20={{t}^{2}}-t-20\]
Now, we have a quadratic equation, we will solve this quadratic equation using the method of the middle term split. This method says equation \[a{{t}^{2}}+bt+c\] will find such a number whose product is the same as \[a\times c\] and their difference or the sum is the same as b. Now, we have,
\[{{t}^{2}}-t-20\]
Here, we have, a = 1, b = – 1 and c = – 20. We have that \[a\times c=1\times -20=-20\] and we can observe that \[4\times -5=-20\] and also 4 – 5 = – 1.
So, we use 4 – 5 = – 1 to split the middle terms using these, we get,
\[{{t}^{2}}-t-20={{t}^{2}}+\left( 4-5 \right)t-20\]
\[\Rightarrow {{t}^{2}}-t-20={{t}^{2}}+4t-5t-20\]
Now, taking the common in the first two terms and the last two terms, we get,
\[\Rightarrow {{t}^{2}}-t-20=t\left( t+4 \right)-5\left( t+4 \right)\]
As (t + 4) is the same, so, we get,
\[\Rightarrow {{t}^{2}}-t-20=\left( t-5 \right)\left( t+4 \right)\]
Now, we substitute t as \[{{x}^{2}},\] so we get,
\[{{x}^{4}}-{{x}^{2}}-20=\left( {{x}^{2}}-5 \right)\left( {{x}^{2}}+4 \right)\]
Now, as we can see that \[{{x}^{2}}+4\] cannot be simplified further and we can factor \[{{x}^{2}}-5\] further. We can write, \[5={{\left( \sqrt{5} \right)}^{2}}\] so we get,
\[\Rightarrow {{x}^{2}}-5={{x}^{2}}-{{\left( \sqrt{5} \right)}^{2}}\]
Now, using \[{{\alpha }^{2}}-{{\beta }^{2}}=\left( \alpha +\beta \right)\left( \alpha -\beta \right)\] and we will consider \[\alpha =x\] and \[\beta =\sqrt{5}\] so we get,
\[{{x}^{2}}-5=\left( {{x}^{2}}-{{\left( \sqrt{5} \right)}^{2}} \right)\]
\[\Rightarrow {{x}^{2}}-5=\left( x+\sqrt{5} \right)\left( x-\sqrt{5} \right)\]
Putting it above, we get,
\[\Rightarrow {{x}^{4}}-{{x}^{2}}-20=\left( {{x}^{2}}+4 \right)\left( x+\sqrt{5} \right)\left( x-\sqrt{5} \right)\]

Note: In the real line, we cannot factor \[{{x}^{2}}+4,\] because if we do it means we have \[{{x}^{2}}+4=0\] which implies \[x=\sqrt{-4}\] which is not true as square root takes only positive values. But if it is in a complex plane, then we can solve it further. We know that \[\sqrt{-1}=i,\] so we can split \[{{x}^{2}}+4\] into linear factor as \[{{x}^{2}}+4=\left( x+2i \right)\left( x-2i \right).\]