How do you factor $4{x^3} + 32$?

Hint: First step is to take out the common factor of 4.
Then use the sum of cubes formula.
And factorize generally to achieve the solution.

Complete Step by Step Solution:
Let us start by writing the given equation:
$ \Rightarrow 4{x^3} + 32$
We will take out the common factor 4 out of $4{x^3}$ we get:
$ \Rightarrow 4\left( {{x^3}} \right) + 32$
Again take out the common factor 4 out of 32 we get:
$ \Rightarrow 4\left( {{x^3}} \right) + 4\left( 8 \right)$
Now, take out the common factor 4 out of the whole above equation we get:
$ \Rightarrow 4\left( {{x^3} + 8} \right)$
Rewrite 8 as ${2^3}$ we get:
$ \Rightarrow 4\left( {{x^3} + {2^3}} \right)$ ……. Let it be eq. $\left( 1 \right)$
Since, both terms are perfect cubes, factor using the sum of cubes formula:
$ \Rightarrow \left( {{a^3} + {b^3}} \right) = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$
Where, the values of the variables are as: $a = x$ and $b = 2$
Now substitute the values of a and b in the sum of cubes formula we will get:
$ \Rightarrow \left( {{x^3} + {2^3}} \right) = \left( {x + 2} \right)\left( {{x^2} - 2x + {2^2}} \right)$ ….. let it be eq. $\left( 2 \right)$
Now, we are going to substitute eq. $\left( 1 \right)$in eq.$\left( 2 \right)$ we will get :
$ \Rightarrow 4\left( {\left( {x + 2} \right)\left( {{x^2} - 2x + {2^2}} \right)} \right)$
Solve ${2^2}$ and put the value of it in the above equation:
$ \Rightarrow 4\left( {\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)} \right)$
Now, by removing unnecessary parenthesis you will get to this:
$ \Rightarrow 4\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)$

Hence, $4\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)$is the required factored form of $4{x^3} + 32$.

Note: Look, we can factor further the term $\left( {{x^2} - 2x + 4} \right)$
The first term is, ${x^2}$ its coefficient is $1$.
The middle term is, $ - 2x$ its coefficient is $ - 2$.
The last term, “the constant”, is$ + 4$.
Now, multiply the coefficient of the first term by the last constant $1 \times 4 = 4$.
Finally, find two factors of 4 whose sum equals the coefficient of the middle term, which is $ - 2$.
As we cannot find any such two factors and therefore we can conclude that trinomial cannot be factored.