How do you factor $8{{x}^{2}}-4x-24$ ?

Hint: When we factorize a quadratic equation $a{{x}^{2}}+bx+c$ we have to find pair of number whose sum is equal to b and product equal to product of a and c . Then we can write bx as the sum of the 2 terms. Here We can split -4x to 12x and -16x to solve this question.

Complete step by step solution:
The given equation is $8{{x}^{2}}-4x-24$ which is a quadratic equation. if we compare the equation to standard quadratic equation $a{{x}^{2}}+bx+c$ then a = 8, b = -4 and c = -24
To factor a quadratic equation, we can find two numbers m and n such that the sum of m and n is equal to b and the product of m and n is $ac$. Then we can split $bx$ to $mx+nx$ then we can factor the equation easily.
In our case ac = -192 and b = -4
So pair of 2 numbers whose product is -192 and sum -4 is ( 12 ,-16)
We can -4x split to 12x – 16x
So $\Rightarrow 8{{x}^{2}}-4x-24=8{{x}^{2}}+12x-16x-24$
Taking 4x common in the first half of the equation and taking -8 common in the second half of the equation.
$\Rightarrow 8{{x}^{2}}-4x-24=4x\left( 2x+3 \right)-8\left( 2x+3 \right)$
Taking 2x + 3 common
$\Rightarrow 8{{x}^{2}}-4x-24=\left( 4x-8 \right)\left( 2x+3 \right)$
We can take 4 common from 4x - 8
$\Rightarrow 8{{x}^{2}}-4x-24=4\left( x-2 \right)\left( 2x+3 \right)$

Note:
While factoring a quadratic equation we can’t always split $bx$ such that their product is equal to ac because sometimes the roots can be irrational numbers. In that case we can find the roots of the equation by formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .