Solve x^3-x^2-22x+40

$$x ^ { 3 } - x ^ { 2 } - 22 x + 40$$

Factor

$$\left(x-4\right)\left(x-2\right)\left(x+5\right)$$

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 40 and q divides the leading coefficient 1. One such root is -5. Factor the polynomial by dividing it by x+5.

$$\left(x+5\right)\left(x^{2}-6x+8\right)$$

Consider x^{2}-6x+8. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=1\times 8=8$$

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 8.

$$-1,-8 -2,-4$$

Calculate the sum for each pair.

$$-1-8=-9 -2-4=-6$$

The solution is the pair that gives sum -6.

$$a=-4 b=-2$$

Rewrite x^{2}-6x+8 as \left(x^{2}-4x\right)+\left(-2x+8\right).

$$\left(x^{2}-4x\right)+\left(-2x+8\right)$$

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

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

Factor out common term x-4 by using distributive property.

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

Rewrite the complete factored expression.

$$\left(x-4\right)\left(x-2\right)\left(x+5\right)$$

Evaluate

$$\left(x-4\right)\left(x-2\right)\left(x+5\right)$$