Solve t^3-5t^2-14t

$$t ^ { 3 } - 5 t ^ { 2 } - 14 t$$

Factor

$$t\left(t-7\right)\left(t+2\right)$$

Solution Steps

Factor out t.

$$t\left(t^{2}-5t-14\right)$$

Consider t^{2}-5t-14. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-14. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-14\right)=-14$$

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 -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=-7 b=2$$

Rewrite t^{2}-5t-14 as \left(t^{2}-7t\right)+\left(2t-14\right).

$$\left(t^{2}-7t\right)+\left(2t-14\right)$$

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

$$t\left(t-7\right)+2\left(t-7\right)$$

Factor out common term t-7 by using distributive property.

$$\left(t-7\right)\left(t+2\right)$$

Rewrite the complete factored expression.

$$t\left(t-7\right)\left(t+2\right)$$

Evaluate

$$t\left(t-7\right)\left(t+2\right)$$