Solve x^3+left(2xright)^2+x=2

$${ x }^{ 3 } + { \left(2x \right) }^{ 2 } +x = 2$$

Solve for x

$$x=\frac{\sqrt{17}-3}{2}\approx 0.561552813$$

$$x=\frac{-\sqrt{17}-3}{2}\approx -3.561552813$$

$$x=-1$$

Solution Steps

Expand \left(2x\right)^{2}.

$$x^{3}+2^{2}x^{2}+x=2$$

Calculate 2 to the power of 2 and get 4.

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

Subtract 2 from both sides.

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

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 1. List all candidates \frac{p}{q}.

$$±2,±1$$

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^{3}+4x^{2}+x-2 by x+1 to get x^{2}+3x-2. Solve the equation where the result equals to 0.

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

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, 3 for b, and -2 for c in the quadratic formula.

$$x=\frac{-3±\sqrt{3^{2}-4\times 1\left(-2\right)}}{2}$$

Do the calculations.

$$x=\frac{-3±\sqrt{17}}{2}$$

Solve the equation x^{2}+3x-2=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{17}-3}{2} x=\frac{\sqrt{17}-3}{2}$$

List all found solutions.

$$x=-1 x=\frac{-\sqrt{17}-3}{2} x=\frac{\sqrt{17}-3}{2}$$