Solve 12r^3+16r^2-60r

$$12 r ^ { 3 } + 16 r ^ { 2 } - 60 r$$

Factor

$$4r\left(3r-5\right)\left(r+3\right)$$

Solution Steps

Factor out 4.

$$4\left(3r^{3}+4r^{2}-15r\right)$$

Consider 3r^{3}+4r^{2}-15r. Factor out r.

$$r\left(3r^{2}+4r-15\right)$$

Consider 3r^{2}+4r-15. Factor the expression by grouping. First, the expression needs to be rewritten as 3r^{2}+ar+br-15. To find a and b, set up a system to be solved.

$$a+b=4 ab=3\left(-15\right)=-45$$

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 -45.

$$-1,45 -3,15 -5,9$$

Calculate the sum for each pair.

$$-1+45=44 -3+15=12 -5+9=4$$

The solution is the pair that gives sum 4.

$$a=-5 b=9$$

Rewrite 3r^{2}+4r-15 as \left(3r^{2}-5r\right)+\left(9r-15\right).

$$\left(3r^{2}-5r\right)+\left(9r-15\right)$$

Factor out r in the first and 3 in the second group.

$$r\left(3r-5\right)+3\left(3r-5\right)$$

Factor out common term 3r-5 by using distributive property.

$$\left(3r-5\right)\left(r+3\right)$$

Rewrite the complete factored expression.

$$4r\left(3r-5\right)\left(r+3\right)$$

Evaluate

$$4r\left(3r-5\right)\left(r+3\right)$$