Solve -27r^3+18r^2+24r

$$- 27 r ^ { 3 } + 18 r ^ { 2 } + 24 r$$

Factor

$$-3r\left(3r-4\right)\left(3r+2\right)$$

Solution Steps

Factor out 3.

$$3\left(-9r^{3}+6r^{2}+8r\right)$$

Consider -9r^{3}+6r^{2}+8r. Factor out r.

$$r\left(-9r^{2}+6r+8\right)$$

Consider -9r^{2}+6r+8. Factor the expression by grouping. First, the expression needs to be rewritten as -9r^{2}+ar+br+8. To find a and b, set up a system to be solved.

$$a+b=6 ab=-9\times 8=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 6.

$$a=12 b=-6$$

Rewrite -9r^{2}+6r+8 as \left(-9r^{2}+12r\right)+\left(-6r+8\right).

$$\left(-9r^{2}+12r\right)+\left(-6r+8\right)$$

Factor out -3r in the first and -2 in the second group.

$$-3r\left(3r-4\right)-2\left(3r-4\right)$$

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

$$\left(3r-4\right)\left(-3r-2\right)$$

Rewrite the complete factored expression.

$$3r\left(3r-4\right)\left(-3r-2\right)$$

Evaluate

$$-3r\left(3r-4\right)\left(3r+2\right)$$