Question

How do you factor ${{x}^{4}}+10{{x}^{2}}+16$ ?

Answer 1

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 $n$ 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:
$\Rightarrow {{x}^{4}}+10{{x}^{2}}+16$
We can rearrange the terms in the expression as:
$\Rightarrow {{\left( {{x}^{2}} \right)}^{2}}+10{{x}^{2}}+16$
Since there is no way to factorize the polynomial directly, let’s consider ${{x}^{2}}=n$. On substituting it in the polynomial, we get:
$\Rightarrow {{n}^{2}}+10n+16$
Now this expression is in the form of $a{{n}^{2}}+bn+c$ therefore, we will solve this as a quadratic equation by splitting the middle term.
Therefore, the equation can be written as:
$\Rightarrow {{n}^{2}}+8n+2n+16$
on taking the common terms, we get:
$\Rightarrow n\left( n+8 \right)+2\left( n+8 \right)$
since the term $\left( n+8 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow \left( n+2 \right)\left( n+8 \right)$
Now on resubstituting the value of $n$ as ${{x}^{2}}$, we get:
$\Rightarrow \left( {{x}^{2}}+2 \right)\left( {{x}^{2}}+8 \right)$, which is the required factorized format of the equation therefore, we can write it as:
$\Rightarrow {{x}^{4}}+10{{x}^{2}}+16=\left( {{x}^{2}}+2 \right)\left( {{x}^{2}}+8 \right)$, which is the required solution.

Note: It is to be noted that the terms $\left( {{x}^{2}}+2 \right)$ and $\left( {{x}^{2}}+8 \right)$are still not linear terms and they cannot be factored further without using complex numbers. If we really want to factorize the equation in linear terms then we can use $\left( {{x}^{2}}+2 \right)=\left( x+i\sqrt{2} \right)\left( x-i\sqrt{2} \right)$ and $\left( {{x}^{2}}+8 \right)=\left( x+i\sqrt{8} \right)\left( x-i\sqrt{8} \right)$ therefore, the factors can be written as ${{x}^{4}}+10{{x}^{2}}+16=\left( x+i\sqrt{2} \right)\left( x-i\sqrt{2} \right)\left( x+i\sqrt{8} \right)\left( x-i\sqrt{8} \right)$ which is the factorized format of writing the equation in linear terms. It is to be noted that the value of $i=\sqrt{-1}$ therefore the value of ${{i}^{2}}=-1$.