How do you factor the expression $15{{x}^{2}}-2x-8$ ?

Hint: We can factorize any quadratic equation $a{{x}^{2}}+bx+c$ by finding 2 numbers such that sum of the 2 numbers is b and product of the 2 numbers is equal to ac. In the equation $15{{x}^{2}}-2x-8$, a = 15, b = -2 and c = -8 . We have to find 2 numbers such that their sum is -2 and the product is equal to – 120.

Complete step by step solution:
The given equation is $15{{x}^{2}}-2x-8$
If we compare it with $a{{x}^{2}}+bx+c$ we get a = 15, b = -2 and c = -8
So, we have to find 2 numbers such that their sum is -2 and the product is – 120. The numbers are 10 and – 12
So, we can write
 $\Rightarrow 15{{x}^{2}}-2x-8=15{{x}^{2}}-12x+10x-8$
We can take 3x common from first 2 terms and 2 common from last 2 terms
$\Rightarrow 15{{x}^{2}}-2x-8=3x\left( 5x-4 \right)+2\left( 5x-4 \right)$
Now we can take 5x – 4 common from the whole equation
$\Rightarrow 15{{x}^{2}}-2x-8=\left( 3x+2 \right)\left( 5x-4 \right)$
$\left( 3x+2 \right)\left( 5x-4 \right)$ is the factored form of the equation $15{{x}^{2}}-2x-8$

Note: If a (x – f)( x – g) is the factored form of the equation $a{{x}^{2}}+bx+c$ , then f and g are 2 roots of the equation $a{{x}^{2}}+bx+c$ . So, if any number d satisfies the equation $a{{x}^{2}}+bx+c$ , then (x- d) is a factor of the equation. If we can not find the factored form of the equation due to irrational root or fractional root then we can use the completing square method to factorize an equation.