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

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

Solve for x

$$x=\sqrt{2}\approx 1.414213562$$

$$x=-\sqrt{2}\approx -1.414213562$$

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

Solution Steps

Subtract 2 from both sides.

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

Reorder the terms.

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -2 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±1,±2,±\frac{1}{2}$$

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 2x^{3}+x^{2}-4x-2 by 2\left(x+\frac{1}{2}\right)=2x+1 to get x^{2}-2. Solve the equation where the result equals to 0.

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

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 -2 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{0±2\sqrt{2}}{2}$$

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

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

List all found solutions.

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