Solve x^4+6x^3-x^2-60x-90=0

$$x ^ { 4 } + 6 x ^ { 3 } - x ^ { 2 } - 60 x - 90 = 0$$

Solve for x

$$x=\sqrt{10}\approx 3.16227766$$

$$x=-\sqrt{10}\approx -3.16227766$$

$$x=-3$$

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 -90 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±90,±45,±30,±18,±15,±10,±9,±6,±5,±3,±2,±1$$

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^{4}+6x^{3}-x^{2}-60x-90 by x+3 to get x^{3}+3x^{2}-10x-30. Solve the equation where the result equals to 0.

$$x^{3}+3x^{2}-10x-30=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -30 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±30,±15,±10,±6,±5,±3,±2,±1$$

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}+3x^{2}-10x-30 by x+3 to get x^{2}-10. Solve the equation where the result equals to 0.

$$x^{2}-10=0$$

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, 0 for b, and -10 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{10}}{2}$$

Solve the equation x^{2}-10=0 when ± is plus and when ± is minus.

$$x=-\sqrt{10} x=\sqrt{10}$$

List all found solutions.

$$x=-3 x=-\sqrt{10} x=\sqrt{10}$$