Solve t^3-19t+30

$$t ^ { 3 } - 19 t + 30$$

Factor

$$\left(t-3\right)\left(t-2\right)\left(t+5\right)$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 30 and q divides the leading coefficient 1. One such root is -5. Factor the polynomial by dividing it by t+5.

$$\left(t+5\right)\left(t^{2}-5t+6\right)$$

Consider t^{2}-5t+6. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 6=6$$

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 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-3 b=-2$$

Rewrite t^{2}-5t+6 as \left(t^{2}-3t\right)+\left(-2t+6\right).

$$\left(t^{2}-3t\right)+\left(-2t+6\right)$$

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

$$t\left(t-3\right)-2\left(t-3\right)$$

Factor out common term t-3 by using distributive property.

$$\left(t-3\right)\left(t-2\right)$$

Rewrite the complete factored expression.

$$\left(t-3\right)\left(t-2\right)\left(t+5\right)$$

Evaluate

$$t^{3}-19t+30$$