Solve r^3+3r^2-54r

$$r ^ { 3 } + 3 r ^ { 2 } - 54 r$$

Factor

$$r\left(r-6\right)\left(r+9\right)$$

Solution Steps

Factor out r.

$$r\left(r^{2}+3r-54\right)$$

Consider r^{2}+3r-54. Factor the expression by grouping. First, the expression needs to be rewritten as r^{2}+ar+br-54. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-54\right)=-54$$

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 -54.

$$-1,54 -2,27 -3,18 -6,9$$

Calculate the sum for each pair.

$$-1+54=53 -2+27=25 -3+18=15 -6+9=3$$

The solution is the pair that gives sum 3.

$$a=-6 b=9$$

Rewrite r^{2}+3r-54 as \left(r^{2}-6r\right)+\left(9r-54\right).

$$\left(r^{2}-6r\right)+\left(9r-54\right)$$

Factor out r in the first and 9 in the second group.

$$r\left(r-6\right)+9\left(r-6\right)$$

Factor out common term r-6 by using distributive property.

$$\left(r-6\right)\left(r+9\right)$$

Rewrite the complete factored expression.

$$r\left(r-6\right)\left(r+9\right)$$

Evaluate

$$r\left(r-6\right)\left(r+9\right)$$