Solve t^3-11t^2+18t

$$t ^ { 3 } - 11 t ^ { 2 } + 18 t$$

Factor

$$t\left(t-9\right)\left(t-2\right)$$

Solution Steps

Factor out t.

$$t\left(t^{2}-11t+18\right)$$

Consider t^{2}-11t+18. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+18. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 18=18$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 18.

$$-1,-18 -2,-9 -3,-6$$

Calculate the sum for each pair.

$$-1-18=-19 -2-9=-11 -3-6=-9$$

The solution is the pair that gives sum -11.

$$a=-9 b=-2$$

Rewrite t^{2}-11t+18 as \left(t^{2}-9t\right)+\left(-2t+18\right).

$$\left(t^{2}-9t\right)+\left(-2t+18\right)$$

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

$$t\left(t-9\right)-2\left(t-9\right)$$

Factor out common term t-9 by using distributive property.

$$\left(t-9\right)\left(t-2\right)$$

Rewrite the complete factored expression.

$$t\left(t-9\right)\left(t-2\right)$$

Evaluate

$$t\left(t-9\right)\left(t-2\right)$$