Factorise: $2{x^2} + 2x - 364$ ?

Hint: Given polynomials of degree $2$. Polynomials of degree $2$ are known as quadratic polynomials. Quadratic polynomials can be factored by the help of splitting the middle term method. In this method, the middle term is split into two terms in such a way that the polynomial remains unchanged.

Complete answer: For factorising the given quadratic polynomial $2{x^2} + 2x - 364$ , we use the splitting the middle term method.
So, we have, $2{x^2} + 2x - 364$
Now, we have to factorise the quadratic polynomial. We can use the splitting of the middle term method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and the product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
First, we take the $2$ common from all the terms.
$ \Rightarrow 2\left( {{x^2} + x - 182} \right)$
Now, we split the middle term $x$ into two terms $14x$ and $ - 13x$ since the product of these terms, $ - 182{x^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $x$.
$ \Rightarrow 2\left( {{x^2} + 14x - 13x - 182} \right)$
Taking x common from the first two terms and $7$ common from the last two terms. We get,
$ \Rightarrow 2\left( {x\left( {x + 14} \right) - 13\left( {x + 14} \right)} \right)$
$ \Rightarrow 2\left( {x - 13} \right)\left( {x + 14} \right)$
So, the factored form of the polynomial $2{x^2} + 2x - 364$ is $2\left( {x - 13} \right)\left( {x + 14} \right)$.

Note:
Splitting of the middle term can be a tedious process at times when the product of the constant term and coefficient of ${x^2}$ is a large number with a large number of divisors. Special care should be taken in such cases. Similar to quadratic polynomials, quadratic equations can also be solved using the factorisation method. Besides factorisation, there are various methods to solve quadratic equations such as completing the square method and using the Quadratic formula.