How do you factor $5{{x}^{3}}-40$ ?

Hint: At first, we take $5$ as common from the entire expression and then express $8$ as ${{2}^{3}}$ . Having done so, the expression becomes of the form ${{a}^{3}}-{{b}^{3}}$ which upon factorization yields $\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ . This is how we solve the problem.

Complete step by step answer:
The given expression is $5{{x}^{3}}-40$ .
Taking $5$ common from the entire expression, we get,
$\Rightarrow 5\times \left( {{x}^{3}}-8 \right)$
Writing $8$ as $2$ raised to the power of $3$ , we get,
$\Rightarrow 5\times \left( {{x}^{3}}-{{2}^{3}} \right)....\text{expression}1$
Now, we know that any expression in the form of ${{a}^{3}}-{{b}^{3}}$ can be factored as
 ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)....equation1$
Comparing this form with $\text{expression}1$ , we get,
$a=x$ and $b=2$
Therefore, factoring $\text{expression}1$ according to $equation1$ , we get
 $\Rightarrow 5\times \left( {{x}^{3}}-{{2}^{3}} \right)=5\times \left( x-2 \right)\left( {{x}^{2}}+\left( 2\times x \right)+{{2}^{2}} \right)$
 $\Rightarrow 5\times \left( {{x}^{3}}-{{2}^{3}} \right)=5\times \left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$
Therefore, we can conclude that the expression $5{{x}^{3}}-40$ can be factorized into $5\times \left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$.

Note:
Remembering the basic factorization formula is a must for these problems. All the perfect squares or cubes should be rewritten in power form for clear understanding and to avoid silly mistakes. This problem can also be solved by a little intuition that if we put $x=2$ in the given expression, it gives $0$ . That means, $x-2$ is a factor of this expression. Therefore, after taking $5$ as common from the entire expression, we divide the expression by $x-2$ . This quotient that we get upon division will be the factor $\left( {{x}^{2}}+2x+4 \right)$ . This method is commonly known as the vanishing factor method.