Solve g(x)=-x^3-x^2+12x

$$g ( x ) = - x ^ { 3 } - x ^ { 2 } + 12 x$$

Factor

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

Solution Steps

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 negative, the negative number has greater absolute value than the positive. 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.

$$a=3 b=-4$$

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

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

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

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

Factor out common term -x+3 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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