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**Hint:**In this question, we are given a quadratic polynomial. It is clear that the first and last terms of this polynomial are not the perfect squares. Therefore, we need to use the method of splitting the middle term to factorize this polynomial.

**Complete step-by-step solution:**

We are given the polynomial $2{x^2} - 9x - 5$.

In this polynomial, we will take $A = 2$, $B = - 9$ and $C = - 5$.

To factorize this polynomial, the middle term should be split in such a way that their product is the same as the product of the coefficients of the first term and the last term.

Therefore, here we need to split $B = - 9$ in such a way that its product will be the product of $A = 2$and $C = - 5$ which is $ - 10$.

So we can take the pair of numbers $ - 10$ and $1$. This is because if we add these two numbers, we get $ - 9$ which is our middle term. And the product of these two numbers, we get $ - 10$ which is the product of the coefficient of the first term and the last term.

Therefore, we can split the middle term as $ - 9x = - 10x + x$

Therefore, we can rewrite the polynomial as

$

2{x^2} - 9x - 5 \\

= 2{x^2} - 10x + x - 5 \\

$

Now we will take $2x$ as common from the first two terms.

$

2{x^2} - 9x - 5 \\

= 2{x^2} - 10x + x - 5 \\

= 2x\left( {x - 5} \right) + 1\left( {x - 5} \right) \\

= \left( {x - 5} \right)\left( {2x + 1} \right) \\

$

Thus, by factoring $2{x^2} - 9x - 5$, we get $\left( {x - 5} \right)\left( {2x + 1} \right)$as our final answer.

**Note:**Here, we have solved the example by factoring the polynomial by splitting its middle term. The general rule for splitting the middle term is that we need to find two numbers whose sum is the same as the middle term. Also, their product should be equal to the product of the coefficients of the first and the last term.

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