Solve P(x)=x^3-x^2-14x+24

$$P ( x ) = x ^ { 3 } - x ^ { 2 } - 14 x + 24$$

Factor

$$\left(x-3\right)\left(x-2\right)\left(x+4\right)$$

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 24 and q divides the leading coefficient 1. One such root is -4. Factor the polynomial by dividing it by x+4.

$$\left(x+4\right)\left(x^{2}-5x+6\right)$$

Consider x^{2}-5x+6. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+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 x^{2}-5x+6 as \left(x^{2}-3x\right)+\left(-2x+6\right).

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

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

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

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

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

Rewrite the complete factored expression.

$$\left(x-3\right)\left(x-2\right)\left(x+4\right)$$

Evaluate

$$\left(x-3\right)\left(x-2\right)\left(x+4\right)$$