Question

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

Answer 1

Hint: In this question we have a polynomial equation which has two terms in it. Since there is no direct way to factorize the equation directly, we will convert the expression into the form of an expansion formula and expand the terms. We will convert the expression into the form of ${{a}^{2}}-{{b}^{2}}$ so that it can be expanded and written in the format of $\left( a+b \right)\left( a-b \right)$.

Complete step by step solution:
We have the given expression as:
$\Rightarrow {{x}^{4}}-4{{x}^{2}}$
Now we can see in the expression that the terms ${{x}^{4}}$ and $4{{x}^{2}}$ are present, which have square roots therefore, we can utilize their square roots to simplify the expression.
We know that ${{x}^{4}}={{\left( {{x}^{2}} \right)}^{2}}$ and $4{{x}^{2}}={{\left( 2x \right)}^{2}}$ therefore, on substituting it in the expression, we get:
$\Rightarrow {{\left( {{x}^{2}} \right)}^{2}}-{{\left( 2x \right)}^{2}}$
Now the above expression is in the form of ${{a}^{2}}-{{b}^{2}}$ therefore, we can expand it using the expansion formula.
On using the formula, we get:
\[\Rightarrow \left( {{x}^{2}}+2x \right)\left( {{x}^{2}}-2x \right)\],
Now the term $x$ is common in both the terms therefore on taking it out as common, we get:
$\Rightarrow \left( x\left( x+2 \right) \right)\left( x\left( x-2 \right) \right)$
On opening the brackets and simplifying the terms, we get:
$\Rightarrow {{x}^{2}}\left( x+2 \right)\left( x-2 \right)$, which is the required solution.

Note: To check whether the solution is correct, we will multiply the factors, if it gives the same term, then the solution is correct:
On multiplying, we get:
$\Rightarrow {{x}^{2}}\left( {{x}^{2}}-4 \right)$
On multiplying the terms, we get:
$\Rightarrow {{x}^{4}}-2{{x}^{2}}$, which is the original expression therefore, the solution is correct.
In the above question we have a polynomial equation of degree $2$. Even though the polynomial equation has a degree $2$, it is not a complete quadratic equation because it has the value of $c$ as $0$.