Solve x^4-5x^3-13x^2+53x+60

$$x ^ { 4 } - 5 x ^ { 3 } - 13 x ^ { 2 } + 53 x + 60$$

Factor

$$\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+3\right)$$

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

$$\left(x-5\right)\left(x^{3}-13x-12\right)$$

Consider x^{3}-13x-12. By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 1. One such root is 4. Factor the polynomial by dividing it by x-4.

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

Consider x^{2}+4x+3. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=4 ab=1\times 3=3$$

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

$$a=1 b=3$$

Rewrite x^{2}+4x+3 as \left(x^{2}+x\right)+\left(3x+3\right).

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

Factor out x in the first and 3 in the second group.

$$x\left(x+1\right)+3\left(x+1\right)$$

Factor out common term x+1 by using distributive property.

$$\left(x+1\right)\left(x+3\right)$$

Rewrite the complete factored expression.

$$\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+3\right)$$

Evaluate

$$\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+3\right)$$