How do you factor $2{{x}^{3}}-3{{x}^{2}}-5x$ ?

Hint: We first convert the cubic expression to a product of a linear and a quadratic expression by taking $x$ common from them. Then, the middle term $-3x$ of the quadratic expression is expressed as $-5x+2x$ . Again taking terms common, we arrive at the simplified form.

Complete step by step answer:
The given expression is
$2{{x}^{3}}-3{{x}^{2}}-5x$
We find that there is a common $x$ in all the three terms. So, for simplification, we need to take this $x$ common. The expression thus becomes,
$\Rightarrow x\left( 2{{x}^{2}}-3x-5 \right)$
Now, we need to simplify the term $\left( 2{{x}^{2}}-3x-5 \right)$ .For this, we need to apply the middle term method. Here, the middle term of the quadratic expression is broken up into a sum of two terms such that their product will be equal to the product of the first and the last terms of the quadratic expression. That is, if $a{{x}^{2}}+bx+c$ , we need to express $b$ as a sum of two terms, say $g$ and $h$ such that, $gh=ac$ .
Thus, applying this for the expression, we get,
$\Rightarrow x\left( 2{{x}^{2}}+2x-5x-5 \right)$
Taking $2x$ common from the first two terms and $-5$ common from the last two terms, we get
$\Rightarrow x\left( 2x\left( x+1 \right)-5\left( x+1 \right) \right)$
Taking $x+1$ common from the two terms, we get,
$\Rightarrow x\left( \left( x+1 \right)\left( 2x-5 \right) \right)$

Therefore, we can conclude that the given expression $2{{x}^{3}}-3{{x}^{2}}-5x$ can be factorized into $x\left( \left( x+1 \right)\left( 2x-5 \right) \right)$

Note: Students may get confused after seeing the cubic expression and may not solve it in that way. So, it's better, if we take $x$ common at first, to make the expression quadratic and easily solvable. The quadratic expression can also be solved by using the vanishing factor method. Here, we see that if we put $x=-1$ the expression gives $0$ . That means, $x+1$ is a factor of the expression. Dividing the expression by $x+1$ gives the other factors.