Solve 16s^3+52s^2-48s

$$16 s ^ { 3 } + 52 s ^ { 2 } - 48 s$$

Factor

$$4s\left(4s-3\right)\left(s+4\right)$$

Solution Steps

Factor out 4.

$$4\left(4s^{3}+13s^{2}-12s\right)$$

Consider 4s^{3}+13s^{2}-12s. Factor out s.

$$s\left(4s^{2}+13s-12\right)$$

Consider 4s^{2}+13s-12. Factor the expression by grouping. First, the expression needs to be rewritten as 4s^{2}+as+bs-12. To find a and b, set up a system to be solved.

$$a+b=13 ab=4\left(-12\right)=-48$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 13.

$$a=-3 b=16$$

Rewrite 4s^{2}+13s-12 as \left(4s^{2}-3s\right)+\left(16s-12\right).

$$\left(4s^{2}-3s\right)+\left(16s-12\right)$$

Factor out s in the first and 4 in the second group.

$$s\left(4s-3\right)+4\left(4s-3\right)$$

Factor out common term 4s-3 by using distributive property.

$$\left(4s-3\right)\left(s+4\right)$$

Rewrite the complete factored expression.

$$4s\left(4s-3\right)\left(s+4\right)$$

Evaluate

$$4s\left(4s-3\right)\left(s+4\right)$$