Question

Factorise\[{\text{: }}2{x^2} - 7x - 15\] .
(A) $\left( {x - 5} \right)\left( {2x + 3} \right)$
(B) $\left( {x - 5} \right)\left( {2x + 1} \right)$
(C) $\left( {x - 6} \right)\left( {2x + 3} \right)$
(D) $\left( {x - 6} \right)\left( {2x + 3} \right)$

Answer 1

Hint: In this question, we have to factorise the given equation. Factorise means we have to make the factors of given equations which can be done by any of the three methods middle term splitting, method of completing squares and discriminant method. This question can be solved by any of these methods and we are going to solve it by a middle term splitting method.

Complete answer:
The given question is to factorise the equation $2{x^2} - 7x - 15{\text{ }}.....................{\text{ (1)}}$
For solving this, we will use a middle term splitting method. Which is :
Firstly, we will multiply the coefficient of ${x^2}$ and the constant term that is
$\Rightarrow$$2 \times 15 = 30$
Now we will make the factors of $30$ and factors of $30$ are $2 \times 3 \times 5$.
Now, we have to arrange these $3$ factors to $2$ factors by multiplication in such a way that if we multiply those $2$ factors we get $30$ (which is the constant term) and when we add or subtract these $2$ factors.
We get $7$ which is the coefficient of $x$. So we arrange $2 \times 3 \times 5$ to $10 \times 3$ because $10$ and $3$, on multiplication gives $30$ and $10$ and $3$. On subtracting gives $7$
So equation (1) becomes $2{x^2} - (10 - 3)x - 15$
In the above equation, we have written
$\Rightarrow$$7$ as $(10 - 3)$
$\therefore $ On multiplying $x$ by both terms ad we get
$\Rightarrow$$2{x^2} - 10x + 3x - 15$
Now we get common $2x$ from first two terms and $3$ from last two terms, we get
$\Rightarrow$$2x(x - 5) + 3(x - 5)$
Now, we get $2$ terms and from these two, we get $(x - 5)$ common out of these two
$\Rightarrow$$(x - 5){\text{ (2x + 3)}}$
Which is option (A)
If means factorization of $2{x^2} - 7x - 15$ is
$\Rightarrow$$\left( {x - 5} \right)\left( {2x + 3} \right)$
And $(x - 5){\text{ \& }}\left( {2x + 3} \right)$ on multiplication gives $2{x^2} - 7x - 15$

So, the correct option is A.

Note: Factorisation of one degree equation or linear equation gives the value of $1$ variable used in the equation. Factorisation of two degree equations gives two factors and factorization of three degree equations gives three factors. Factorisation of three degree equation can be evaluated by deducing three degree equation to the multiplication of one & two degree equation and on further solving two degree equation we finally get $3$ factors of cubic equation.