Solve x^3+3x^2-10x-24

$$x ^ { 3 } + 3 x ^ { 2 } - 10 x - 24$$

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.

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

Consider x^{2}-x-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.

$$a+b=-1 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite x^{2}-x-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

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