Solve t^3-7t+6

$$t ^ { 3 } - 7 t + 6$$

Factor

$$\left(t-2\right)\left(t-1\right)\left(t+3\right)$$

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 6 and q divides the leading coefficient 1. One such root is -3. Factor the polynomial by dividing it by t+3.

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

Consider t^{2}-3t+2. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt+2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

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

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

Factor out t in the first and -1 in the second group.

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

Factor out common term t-2 by using distributive property.

$$\left(t-2\right)\left(t-1\right)$$

Rewrite the complete factored expression.

$$\left(t-2\right)\left(t-1\right)\left(t+3\right)$$

Evaluate

$$t^{3}-7t+6$$