Solve x^4-10x^3-62x^2-78x-27=0

$${ x }^{ 4 } -10 { x }^{ 3 } -62 { x }^{ 2 } -78x-27 = 0$$

Solve for x

$$x=\sqrt{58}+7\approx 14.615773106$$

$$x=7-\sqrt{58}\approx -0.615773106$$

$$x=-1$$

$$x=-3$$

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 -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-10x^{3}-62x^{2}-78x-27 by x+1 to get x^{3}-11x^{2}-51x-27. Solve the equation where the result equals to 0.

$$x^{3}-11x^{2}-51x-27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -27 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±27,±9,±3,±1$$

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}-11x^{2}-51x-27 by x+3 to get x^{2}-14x-9. Solve the equation where the result equals to 0.

$$x^{2}-14x-9=0$$

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, -14 for b, and -9 for c in the quadratic formula.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$x=\frac{14±2\sqrt{58}}{2}$$

Solve the equation x^{2}-14x-9=0 when ± is plus and when ± is minus.

$$x=7-\sqrt{58} x=\sqrt{58}+7$$

List all found solutions.

$$x=-1 x=-3 x=7-\sqrt{58} x=\sqrt{58}+7$$