How do you factor the expression $16{{b}^{2}}+60b-100$ ?

Hint: The given equation $16{{b}^{2}}+60b-100$ is a quadratic equation. To factorize a quadratic equation $a{{x}^{2}}+bx+c$ we have to write bx as mx + nx such that product of m and n is equal to product of ac. In the equation $16{{b}^{2}}+60b-100$ we have to write 60b = mb + nb such that mn is equal to – 1600. We can evaluate m is 80 and n is – 20.

Complete step by step solution:
The given equation is $16{{b}^{2}}+60b-100$
If we compare it with $a{{x}^{2}}+bx+c$ we get a = 16, b = 60 and c = - 100
The value of ac is equal to – 1600
So, we have to find 2 numbers such that their sum is 60 and product is – 1600. The numbers are 80 and – 20
So, we can write
 $\Rightarrow 16{{b}^{2}}+60b-100=16{{b}^{2}}+80b-20b-100$
We can take 16b common from first 2 terms and – 20 common from last 2 terms
$\Rightarrow 16{{b}^{2}}+60b-100=16b\left( b+5 \right)-20\left( b+5 \right)$
Now we can take b + 5 common from the whole equation
$\Rightarrow 16{{b}^{2}}+60b-100=\left( 16b-20 \right)\left( b+5 \right)$
Now we can take 4 common from 16b – 20
$\Rightarrow 16{{b}^{2}}+60b-100=4\left( 4b-5 \right)\left( b+5 \right)$

$4\left( 4b-5 \right)\left( b+5 \right)$ is the factored form of the equation $16{{b}^{2}}+60b-100$.

Note: The graph of all quadratic equations is a parabola whose axis is always parallel to y axis. In quadratic equation $a{{x}^{2}}+bx+c$ , if a is greater than 0 then it will be upward parabola and if a is less than 0 then graph will be downward parabola. If the discriminant (${{b}^{2}}-4ac$ ) is less than 0 then the parabola will never touch the x axis.