Solve 4x^3+2x^2-2x-1=0

$$4 x ^ { 3 } + 2 x ^ { 2 } - 2 x - 1 = 0$$

Solve for x

$$x=\frac{\sqrt{2}}{2}\approx 0.707106781$$

$$x=-\frac{\sqrt{2}}{2}\approx -0.707106781$$

$$x=-\frac{1}{2}=-0.5$$

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

$$±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$x=-\frac{1}{2}$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}+2x^{2}-2x-1 by 2\left(x+\frac{1}{2}\right)=2x+1 to get 2x^{2}-1. Solve the equation where the result equals to 0.

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

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 2 for a, 0 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{0±2\sqrt{2}}{4}$$

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

$$x=-\frac{\sqrt{2}}{2} x=\frac{\sqrt{2}}{2}$$

List all found solutions.

$$x=-\frac{1}{2} x=-\frac{\sqrt{2}}{2} x=\frac{\sqrt{2}}{2}$$