Question

How do you factor ${{x}^{4}}+6{{x}^{2}}-7$ completely?

Answer 1

Hint: In this question we have a polynomial equation which cannot be factorized directly. We know that the general form of a quadratic equation is $a{{x}^{2}}+bx+c$ therefore, in this question we will consider the term ${{x}^{2}}$ as some variable $c$ 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:
$\Rightarrow {{x}^{4}}+6{{x}^{2}}-7$
We can rearrange the terms in the expression as:
$\Rightarrow {{\left( {{x}^{2}} \right)}^{2}}+6{{x}^{2}}-7$
Since there is no way to factorize the polynomial directly, let's consider ${{x}^{2}}=c$. On substituting it in the polynomial, we get:
$\Rightarrow {{c}^{2}}+6c-7$
Now this expression is in the form of $a{{x}^{2}}+bx+c$ therefore, we will solve this as a quadratic equation by splitting the middle term.
Therefore, the equation can be written as:
$\Rightarrow {{c}^{2}}+7c-c-7$
on taking the common terms, we get:
$\Rightarrow c\left( c+7 \right)-1\left( c+7 \right)$
since the term $\left( c+7 \right)$ is common in both the terms, we can take it out as common and write the equation as:
$\Rightarrow \left( c+7 \right)\left( c-1 \right)$
Now on substituting the value of $c$ as ${{x}^{2}}$, we get:
$\Rightarrow \left( {{x}^{2}}+7 \right)\left( {{x}^{2}}-1 \right)$
Since the term $\left( {{x}^{2}}-1 \right)$ is in the form of $\left( {{a}^{2}}-{{b}^{2}} \right)$ we can expand it further as:
$\Rightarrow \left( {{x}^{2}}+7 \right)\left( x-1 \right)\left( x+1 \right)$, which is the required solution.

Note: It is to be remembered that a polynomial equation is a combination of variables and their coefficients, the equation given above is a $4th$ degree polynomial equation.
It is not necessary that all the quadratic equations would have roots which are integer numbers or real numbers therefore quadratic formula is used to solve these types of questions, the quadratic formula is:
$({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the quadratic equation.