Solve w^3-4w^2-12w

$$w ^ { 3 } - 4 w ^ { 2 } - 12 w$$

Factor

$$w\left(w-6\right)\left(w+2\right)$$

Solution Steps

Factor out w.

$$w\left(w^{2}-4w-12\right)$$

Consider w^{2}-4w-12. Factor the expression by grouping. First, the expression needs to be rewritten as w^{2}+aw+bw-12. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-12\right)=-12$$

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 -4.

$$a=-6 b=2$$

Rewrite w^{2}-4w-12 as \left(w^{2}-6w\right)+\left(2w-12\right).

$$\left(w^{2}-6w\right)+\left(2w-12\right)$$

Factor out w in the first and 2 in the second group.

$$w\left(w-6\right)+2\left(w-6\right)$$

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

$$\left(w-6\right)\left(w+2\right)$$

Rewrite the complete factored expression.

$$w\left(w-6\right)\left(w+2\right)$$

Evaluate

$$w\left(w-6\right)\left(w+2\right)$$