Solve x^4-27x^2+14x+120=0

$$x ^ { 4 } - 27 x ^ { 2 } + 14 x + 120 = 0$$

Solve for x

$$x=3$$

$$x=4$$

$$x=-2$$

$$x=-5$$

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. List all candidates \frac{p}{q}.

$$±120,±60,±40,±30,±24,±20,±15,±12,±10,±8,±6,±5,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-27x^{2}+14x+120 by x+2 to get x^{3}-2x^{2}-23x+60. Solve the equation where the result equals to 0.

$$x^{3}-2x^{2}-23x+60=0$$

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. List all candidates \frac{p}{q}.

$$±60,±30,±20,±15,±12,±10,±6,±5,±4,±3,±2,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-2x^{2}-23x+60 by x-3 to get x^{2}+x-20. Solve the equation where the result equals to 0.

$$x^{2}+x-20=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 1 for b, and -20 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 1\left(-20\right)}}{2}$$

Do the calculations.

$$x=\frac{-1±9}{2}$$

Solve the equation x^{2}+x-20=0 when ± is plus and when ± is minus.

$$x=-5 x=4$$

List all found solutions.

$$x=-2 x=3 x=-5 x=4$$