Solve t^3-4t^2-12t

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

Factor

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

Solution Steps

Factor out t.

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

Consider t^{2}-4t-12. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-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 t^{2}-4t-12 as \left(t^{2}-6t\right)+\left(2t-12\right).

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

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

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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