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

Hint: 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: -
\[\begin{align}
  & \Rightarrow \left( x-2 \right)\left( {{x}^{2}}+{{2}^{2}}+2x \right)=0 \\
 & \Rightarrow \left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)=0 \\
\end{align}\]
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) = \[{{b}^{2}}-4ac\]
\[\begin{align}
  & \Rightarrow D={{\left( 2 \right)}^{2}}-4\left( 1 \right)\left( 4 \right) \\
 & \Rightarrow D=4-16 \\
 & \Rightarrow D=-12 \\
 & \Rightarrow D<0 \\
\end{align}\]
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.