Solve x^4-10x^3+28x^2-15x-4=0

$$x ^ { 4 } - 10 x ^ { 3 } + 28 x ^ { 2 } - 15 x - 4 = 0$$

Solve for x

$$x=\frac{5-\sqrt{29}}{2}\approx -0.192582404$$

$$x=4$$

$$x = \frac{\sqrt{29} + 5}{2} \approx 5.192582404$$

$$x=1$$

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 -4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$x=1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-10x^{3}+28x^{2}-15x-4 by x-1 to get x^{3}-9x^{2}+19x+4. Solve the equation where the result equals to 0.

$$x^{3}-9x^{2}+19x+4=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 4 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±4,±2,±1$$

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.

$$x=4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+19x+4 by x-4 to get x^{2}-5x-1. Solve the equation where the result equals to 0.

$$x^{2}-5x-1=0$$

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, -5 for b, and -1 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\left(-1\right)}}{2}$$

Do the calculations.

$$x=\frac{5±\sqrt{29}}{2}$$

Solve the equation x^{2}-5x-1=0 when ± is plus and when ± is minus.

$$x=\frac{5-\sqrt{29}}{2} x=\frac{\sqrt{29}+5}{2}$$

List all found solutions.

$$x=1 x=4 x=\frac{5-\sqrt{29}}{2} x=\frac{\sqrt{29}+5}{2}$$