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

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

Solve for x

$$x=-1$$

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

Solution Steps

Subtract 3 from both sides.

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

Subtract 3 from 2 to get -1.

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

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

$$±\frac{1}{2},±1$$

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.

$$x=-1$$

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

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

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

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

Do the calculations.

$$x=\frac{-1±3}{4}$$

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

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

List all found solutions.

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