Solve x^3+x^2-14x-24

$$x ^ { 3 } + x ^ { 2 } - 14 x - 24$$

Factor

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

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 -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 positive, a and b are both positive. 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.

$$a=2 b=3$$

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

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

Factor out x in the first and 3 in the second group.

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

Factor out common term x+2 by using distributive property.

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

Rewrite the complete factored expression.

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

Evaluate

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