Solve 3x^3+5x^2=3x+2

$$3 x ^ { 3 } + 5 x ^ { 2 } = 3 x + 2$$

Solve for x

$$x=\frac{\sqrt{13}+1}{6}\approx 0.767591879$$

$$x=\frac{1-\sqrt{13}}{6}\approx -0.434258546$$

$$x=-2$$

Solution Steps

Subtract 3x from both sides.

$$3x^{3}+5x^{2}-3x=2$$

Subtract 2 from both sides.

$$3x^{3}+5x^{2}-3x-2=0$$

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

$$±\frac{2}{3},±2,±\frac{1}{3},±1$$

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.

$$x=-2$$

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

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

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

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

Do the calculations.

$$x=\frac{1±\sqrt{13}}{6}$$

Solve the equation 3x^{2}-x-1=0 when ± is plus and when ± is minus.

$$x=\frac{1-\sqrt{13}}{6} x=\frac{\sqrt{13}+1}{6}$$

List all found solutions.

$$x=-2 x=\frac{1-\sqrt{13}}{6} x=\frac{\sqrt{13}+1}{6}$$