Solve 12r^3+39r^2-36r

$$12 r ^ { 3 } + 39 r ^ { 2 } - 36 r$$

Factor

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

Solution Steps

Factor out 3.

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

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

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

Consider 4r^{2}+13r-12. Factor the expression by grouping. First, the expression needs to be rewritten as 4r^{2}+ar+br-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 4r^{2}+13r-12 as \left(4r^{2}-3r\right)+\left(16r-12\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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