How do you factorize $14{x^2} - x - 3$ ?

Hint:In the above problem you were asked to factorize $14{x^2} - x - 3$. You can solve this problem using the middle term factorization. In the middle term factorization, we rewrite the middle term by adding or subtracting them in such a way that we can take common and find the factors. So let us see how we can solve this problem.

Complete Step by Step Solution:
In the given statement we have to factorize $14{x^2} - x - 3$. We will use middle term factorization to solve this problem.
 $= 14{x^2} - x - 3$
-1 can also be written as – 7 + 6
 $= 14{x^2} - 7x + 6x - 3$
After taking common we will get two pairs of factors
 $= 7x(2x - 1) + 3(2x - 1)$
 $= (2x - 1)(7x + 3)$

Therefore, factors of $14{x^2} - x - 3$ is (2x - 1) (7x + 3).

Additional Information:
In the above solution, we used the formula of middle term factorization. A quadratic equation $a{x^2} + bx + c$ can be solved with the same formula. The quadratic equation is of different types: it can be a standard quadratic equation, it can be factored or it can be solved with the quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Note:
In the above problem, the equation was given but it did not equal 0. It means that we do not have to find the value of x rather we have to check for the factors and find only the factors which will be our answer.