Solve x^4+6x^3-5x^2-42x+40

$$x ^ { 4 } + 6 x ^ { 3 } - 5 x ^ { 2 } - 42 x + 40$$

Factor

$$\left(x-2\right)\left(x-1\right)\left(x+4\right)\left(x+5\right)$$

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.

$$\left(x+5\right)\left(x^{3}+x^{2}-10x+8\right)$$

Consider x^{3}+x^{2}-10x+8. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 8 and q divides the leading coefficient 1. One such root is -4. Factor the polynomial by dividing it by x+4.

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

Consider x^{2}-3x+2. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-2 b=-1$$

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

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

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

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

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

$$\left(x-2\right)\left(x-1\right)$$

Rewrite the complete factored expression.

$$\left(x-2\right)\left(x-1\right)\left(x+4\right)\left(x+5\right)$$

Evaluate

$$\left(x-2\right)\left(x-1\right)\left(x+4\right)\left(x+5\right)$$