Solve -3x^3+3x^2+36x

$$- 3 x ^ { 3 } + 3 x ^ { 2 } + 36 x$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(-x^{3}+x^{2}+12x\right)$$

Consider -x^{3}+x^{2}+12x. Factor out x.

$$x\left(-x^{2}+x+12\right)$$

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

$$a+b=1 ab=-12=-12$$

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 -12.

$$-1,12 -2,6 -3,4$$

Calculate the sum for each pair.

$$-1+12=11 -2+6=4 -3+4=1$$

The solution is the pair that gives sum 1.

$$a=4 b=-3$$

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

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

Factor out -x in the first and -3 in the second group.

$$-x\left(x-4\right)-3\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(-x-3\right)$$

Rewrite the complete factored expression.

$$3x\left(x-4\right)\left(-x-3\right)$$

Evaluate

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