Solve x^3-23x^2+142x-120

$$x ^ { 3 } - 23 x ^ { 2 } + 142 x - 120$$

Factor

$$\left(x-12\right)\left(x-10\right)\left(x-1\right)$$

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 -120 and q divides the leading coefficient 1. One such root is 12. Factor the polynomial by dividing it by x-12.

$$\left(x-12\right)\left(x^{2}-11x+10\right)$$

Consider x^{2}-11x+10. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=-11 ab=1\times 10=10$$

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 10.

$$-1,-10 -2,-5$$

Calculate the sum for each pair.

$$-1-10=-11 -2-5=-7$$

The solution is the pair that gives sum -11.

$$a=-10 b=-1$$

Rewrite x^{2}-11x+10 as \left(x^{2}-10x\right)+\left(-x+10\right).

$$\left(x^{2}-10x\right)+\left(-x+10\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(x-10\right)-\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-1\right)$$

Rewrite the complete factored expression.

$$\left(x-12\right)\left(x-10\right)\left(x-1\right)$$

Evaluate

$$\left(x-12\right)\left(x-10\right)\left(x-1\right)$$