How do you find the roots of ${x^3} - 12x + 16 = 0$?

Hint: We will first find one root of the given function ${x^3} - 12x + 16 = 0$ using hit and trial method. Then, we will make one factor using it and divide the given equation by it and get the required factors and roots.

Complete step by step answer:
We are given that we are required to find the roots of ${x^3} - 12x + 16 = 0$.
Let us say $f(x) = {x^3} - 12x + 16$.
Putting $x = 2$ in the given equation $f(x) = {x^3} - 12x + 16$.
We will then obtain: $f(2) = {2^3} - 12 \times 2 + 16$
Simplifying the calculations, we will then obtain the following:-
$ \Rightarrow f(2) = 8 - 24 + 16 = 0$
Therefore, $x = 2$ which means $\left( {x - 2} \right)$ is a factor of $f(x)$.
Dividing the given equation $f(x) = {x^3} - 12x + 16$ by $\left( {x - 2} \right)$, we will then obtain the following equation:-
$ \Rightarrow x - 2)\overline {{x^3} - 12x + 16} $
Multiplying the divisor by ${x^2}$, we will then obtain the following equation with us:-
                  ${x^2}$
$ \Rightarrow x - 2)\overline {{x^3} - 12x + 16} $
                 ${x^3} - 2{x^2}$
                 $\overline {2{x^2} - 12x + 16} $
Multiplying the divisor by 2x now, we will then obtain the following equation:-
                  ${x^2} + 2x$
$ \Rightarrow x - 2)\overline {{x^3} - 12x + 16} $
                 ${x^3} - 2{x^2}$
                 $\overline {2{x^2} - 12x + 16} $
                 $2{x^2} - 4x$
                 $\overline { - 8x + 16} $
Multiplying the divisor by – 8 now, we will then obtain the following equation:-
                  ${x^2} + 2x - 8$
$ \Rightarrow x - 2)\overline {{x^3} - 12x + 16} $
                 ${x^3} - 2{x^2}$
                 $\overline {2{x^2} - 12x + 16} $
                 $2{x^2} - 4x$
                 $\overline { - 8x + 16} $
                 $ - 8x + 16$
                  _______
                        0
Thus, we can write the given polynomial ${x^3} - 12x + 16 = 0$ as follows:-
$ \Rightarrow x{x^3} - 12x + 16 = 0 = \left( {x - 2} \right)\left( {{x^2} + 2x - 8} \right)$
We can write the latter equation as follows:-
$ \Rightarrow x{x^3} - 12x + 16 = 0 = \left( {x - 2} \right)\left( {{x^2} - 2x + 4x - 8} \right)$
Taking x common from the first two terms in the latter factor, we will then obtain the following equation with us:-
$ \Rightarrow x{x^3} - 12x + 16 = 0 = \left( {x - 2} \right)\left\{ {x\left( {x - 2} \right) + 4x - 8} \right\}$
Taking 4 common from the last two terms in the latter factor, we will then obtain the following equation with us:-
$ \Rightarrow x{x^3} - 12x + 16 = 0 = \left( {x - 2} \right)\left\{ {x\left( {x - 2} \right) + 4\left( {x - 2} \right)} \right\}$
Taking $\left( {x - 2} \right)$ common from the last two terms in the latter factor, we will then obtain the following equation with us:-
$ \Rightarrow x{x^3} - 12x + 16 = 0 = {\left( {x - 2} \right)^2}\left( {x + 4} \right)$
Therefore, the roots are 2, 2 and – 4.

Note:
The students must notice that we have an alternate way of factoring the quadratic equation involved in it as well. The alternate way is as follows:-
The given equation is ${x^2} + 2x - 8$.
Using the quadratic formula given by if the equation is given by $a{x^2} + bx + c = 0$, its roots are given by the following equation:-
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Thus, we have the roots of ${x^2} + 2x - 8$ given by:
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 32} }}{2}$
Hence, the roots are - 4 and 2.