Solve x^3+19x^2+118x+240

$$x ^ { 3 } + 19 x ^ { 2 } + 118 x + 240$$

Factor

$$\left(x+5\right)\left(x+6\right)\left(x+8\right)$$

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 240 and q divides the leading coefficient 1. One such root is -8. Factor the polynomial by dividing it by x+8.

$$\left(x+8\right)\left(x^{2}+11x+30\right)$$

Consider x^{2}+11x+30. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=11 ab=1\times 30=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 11.

$$a=5 b=6$$

Rewrite x^{2}+11x+30 as \left(x^{2}+5x\right)+\left(6x+30\right).

$$\left(x^{2}+5x\right)+\left(6x+30\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(x+5\right)+6\left(x+5\right)$$

Factor out common term x+5 by using distributive property.

$$\left(x+5\right)\left(x+6\right)$$

Rewrite the complete factored expression.

$$\left(x+5\right)\left(x+6\right)\left(x+8\right)$$

Evaluate

$$\left(x+5\right)\left(x+6\right)\left(x+8\right)$$