How do you factor the expression $16{{x}^{2}}-8x-15$ ?

Hint: We can factorize any quadratic equation $a{{x}^{2}}+bx+c$ by writing bx as sum of 2 terms such that the sum of the coefficient of the 2 terms is b and product is ac. In this case the sum of coefficients should be - 8 and product should be – 240. We can find the numbers -20 and 12.

Complete step by step solution:
The given equation is $16{{x}^{2}}-8x-15$
If we compare it with $a{{x}^{2}}+bx+c$ we get a = 16, b = - 8 and c = - 15
So the value of ac is equal to -240
So, we have to find 2 numbers such that their sum is - 8 and product is – 240 . The numbers are 8 and – 20
So, we can write
 $\Rightarrow 16{{x}^{2}}-8x-15=16{{x}^{2}}-20x+12x-15$
We can take 4x common from first 2 terms and 3 common from last 2 terms
$\Rightarrow 16{{x}^{2}}-8x-15=4x\left( 4x-5 \right)+3\left( 4x-5 \right)$
Now we can take 5x – 4 common from the whole equation
$\Rightarrow 16{{x}^{2}}-8x-15=\left( 4x+3 \right)\left( 4x-5 \right)$
$\left( 4x+3 \right)\left( 4x-5 \right)$ is the factored form of the equation $16{{x}^{2}}-8x-15$

Note: Sometimes in some equations we can not split the term bx due to irrational roots or decimal roots. In that case we can factorize the equation by complete square method. The discriminant of the quadratic equation $a{{x}^{2}}+bx+c$is ${{b}^{2}}-4ac$. If ${{b}^{2}}-4ac$ is less than 0, we can not find any real roots of the quadratic equation. 2 roots will be imaginary. In that case we can not factorize the equation.