Solve 4x^3-12x^2-16x=

$$4 x ^ { 3 } - 12 x ^ { 2 } - 16 x =$$

Factor

$$4x\left(x-4\right)\left(x+1\right)$$

Solution Steps

Factor out 4.

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

Consider x^{3}-3x^{2}-4x. Factor out x.

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

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

$$a+b=-3 ab=1\left(-4\right)=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=-4 b=1$$

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

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

Factor out x in x^{2}-4x.

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

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

$$\left(x-4\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$4x\left(x-4\right)\left(x+1\right)$$

Evaluate

$$4x\left(x-4\right)\left(x+1\right)$$