Solve p^3-7p-6

$$p ^ { 3 } - 7 p - 6$$

Factor

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

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 p-3.

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

Consider p^{2}+3p+2. Factor the expression by grouping. First, the expression needs to be rewritten as p^{2}+ap+bp+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 positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

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

$$\left(p^{2}+p\right)+\left(2p+2\right)$$

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

$$p\left(p+1\right)+2\left(p+1\right)$$

Factor out common term p+1 by using distributive property.

$$\left(p+1\right)\left(p+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

$$p^{3}-7p-6$$