Question

How do you factor the quadratic equation $4{{x}^{3}}-32=0$?

Answer 1

Divide both the sides of the given equation by 4 to get the coefficient of ${{x}^{3}}$ equal to 1. Now, write the obtained equation in the form ${{a}^{3}}-{{b}^{3}}$ and apply the formula: - ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$ to get the factored form. Now, check if the obtained quadratic polynomial can be further factored or not by finding its discriminant value. If $D\ge 0$ then it can be factored and if D < 0 then it cannot be factored in real numbers.

Complete step-by-step solution:
Here, we have been provided with the cubic equation $4{{x}^{3}}-32=0$ and we are asked to factor it.
Now, dividing both sides of the equation with 4 to make the coefficient of ${{x}^{3}}$ equal to 1 and using the fact that 0 divided by any non – zero number equals 0, we get,
$\Rightarrow {{x}^{3}}-8=0$
We can write 8 as ${{2}^{3}}$, so using this conversion, we have,
$\Rightarrow {{x}^{3}}-{{2}^{3}}=0$
Clearly, we can see that the above equation is of the form ${{a}^{3}}-{{b}^{3}}$ whose factored form is given by the formula: - ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. So, using this relation we can write: -

& \Rightarrow \left( x-2 \right)\left( {{x}^{2}}+{{2}^{2}}+2x \right)=0 \\\ & \Rightarrow \left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)=0 \\\ \end{aligned}$$ Now, let us check if the quadratic polynomial $$\left( {{x}^{2}}+2x+4 \right)$$ can be further factored or not. To do so, we need to find its discriminant value. If the discriminant value is greater than or equal to 0 then it can be factored and if the discriminant value is less than 0 then it cannot be factored in real numbers. Let us check. Assuming a, b and c as the coefficient of $${{x}^{2}}$$, coefficient of x and the constant term, we get, $$\Rightarrow $$ Discriminant (D) = $${{b}^{2}}-4ac$$ $$\begin{aligned} & \Rightarrow D={{\left( 2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right) \\\ & \Rightarrow D=4-16 \\\ & \Rightarrow D=-12 \\\ & \Rightarrow D<0 \\\ \end{aligned}$$ Therefore, we cannot factor $$\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$$ further. **Hence, $$\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)=0$$ is our required factored form and answer.** **Note:** One may note that here we are not asked to find the value of x that is why we stopped at the factored form only. You can determine all the roots of the equation by substituting each term equal to 0 one – by – one. Note that we can factor $${{x}^{2}}+2x+4$$, but the factors will be complex numbers like: - $$-1+\sqrt{3}i$$ and $$-1-\sqrt{3}i$$. As you can see that here the coefficients of $${{x}^{2}}$$ and x were 0 and that is why the formula: - $${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$$ was applicable. If the equation were of the form $$a{{x}^{3}}+b{{x}^{2}}+cx+d=0$$ where $$\left( a,b,c,d \right)\ne 0$$ , then in this case we would have found one root by the hit and trial method and the other two roots by the middle term split method.