Question

How do you factor ${{x}^{4}}-8x$?

Answer 1

In this type of question, during solving we should remember a formula that is ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. In this question, first will take out the common factor x. After that, we will factor the remaining equation using the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.

Complete answer:
Let us solve this question.
It is asked in the question how to factor ${{x}^{4}}-8x$.
So, we will factorize the equation ${{x}^{4}}-8x$.
Hence, for the factorization of the given equation ${{x}^{4}}-8x$, first we will take out the common factor x, because it can be seen that x is a factor in both terms that is ${{x}^{4}}$ and $8x$.
We can write the given equation as
${{x}^{4}}-8x=x\times x\times x\times x-8\times x$
As x is common in the above equation. so, we can write the above equation as
${{x}^{4}}-8x=x\times \left( x\times x\times x-8 \right)$
${{x}^{4}}-8x=x\times \left( {{x}^{3}}-8 \right)$
${{x}^{4}}-8x=x\left( {{x}^{3}}-8 \right)$
As we know that ${{2}^{3}}=8$.
So, the equation can also be written as
$\Rightarrow {{x}^{4}}-8x=x\left( {{x}^{3}}-{{2}^{3}} \right)$
Now, we can see that ${{x}^{3}}-{{2}^{3}}$ is in the form of ${{a}^{3}}-{{b}^{3}}$ where a is x and b is 2.
And we know that there is a formula for ${{a}^{3}}-{{b}^{3}}$ that is ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.
So, we can write the equation ${{x}^{3}}-{{2}^{3}}$ as
${{x}^{3}}-{{2}^{3}}=\left( x-2 \right)\left( {{x}^{2}}+{{2}^{2}}+2x \right)=\left( x-2 \right)\left( {{x}^{2}}+4+2x \right)$
Hence,
$\Rightarrow {{x}^{4}}-8x=x\left( {{x}^{3}}-{{2}^{3}} \right)=x\left( x-2 \right)\left( {{x}^{2}}+4+2x \right)$
$\Rightarrow {{x}^{4}}-8x=x\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$
Now, we cannot do the factorization of ${{x}^{2}}+2x+4$, because it has no roots and no factors.
Now, we have done the factorization of ${{x}^{4}}-8x$.
So, the factors of ${{x}^{4}}-8x$ are $x$, $x-2$, and ${{x}^{2}}+2x+4$.

Note: In this type of question, whenever we are solving this, we should remember cubes of some specific numbers like ${{1}^{3}}=1$, ${{2}^{3}}=8$, etc.
Don’t forget to check factors of ${{x}^{2}}+2x+4$.
The factors can be found by finding the real roots of the equation.
If the equation has real roots, then the equation will have factors.
For checking the real roots of the equation, we check if the discriminant is positive or zero. If the discriminant of the equation is negative, then it will have no roots.
As we know that the discriminant of any general quadratic equation $a{{x}^{2}}+bx+c=0$ is $D={{b}^{2}}-4ac$
Hence, discriminant of equation ${{x}^{2}}+2x+4$ is
$D={{2}^{2}}-4\times 1\times 4=4-16=-12$
As it is negative, so the equation will have no roots. And if it has no roots, then it will have no factors.