Question

How do you factor $12{x^2} + 13x - 14$ ?

Answer 1

Hint: This is an algebraic expression of degree two. So, it is a quadratic equation. The factors of the expression can be found using factoring or by using the quadratic formula. Since, it is a quadratic expression it will have two factors at most.

Complete step-by-step answer:
General form of a quadratic equation is $a{x^2} + bx + c$. Here the given equation is in the required standard form.
$12{x^2} + 13x - 14$ -------------- (1)
Next, we will compare this equation with the standard form and find the values of $a$, $b$and $c$.
So, we get, $a = 12,{\text{ }}b = 13{\text{ and }}c = - 14$.
Next, we will equate this expression to zero.
$12{x^2} + 13x - 14 = 0$
Now, we will use the quadratic formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ for finding the roots of the equation and use the roots to find the factors. We will proceed by calculating the determinant of the equation first.
$D = {b^2} - 4ac$
$ \Rightarrow D = {13^2} - 4(12)( - 14)$
$ \Rightarrow D = 169 + 672$
$ \Rightarrow D = 841$
So, the equation will have two distinct real roots.
Now, using the formula, we have
\[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
\[ \Rightarrow x = \dfrac{{ - 13 \pm \sqrt {841} }}{{2(12)}}\]
\[ \Rightarrow x = \dfrac{{ - 13 \pm 29}}{{24}}\]
\[ \Rightarrow x = \dfrac{{ - 13 + 29}}{{24}}{\text{ and }}x = \dfrac{{ - 13 - 29}}{{24}}\]
\[ \Rightarrow x = \dfrac{{16}}{{24}}{\text{ = }}\dfrac{2}{3}{\text{ and }}x = \dfrac{{ - 42}}{{24}} = \dfrac{{ - 7}}{4}\]
So, the factors of the equation are \[x - \dfrac{2}{3}{\text{ and }}x + \dfrac{7}{4}\].
Hence, $12{x^2} + 13x - 14 = \left( {3x - 2} \right)\left( {4x + 7} \right)$.
So, the correct answer is “$\left( {3x - 2} \right)\left( {4x + 7} \right)$”.

Note: Be careful in calculations when using the quadratic formula. Here, factoring would be tedious. So, it is better to actually go with the quadratic formula when the numbers whose product gives $ac$and sum is equal to $b$.