Solve 2x^3+3x^2=8x+12

$$2 x ^ { 3 } + 3 x ^ { 2 } = 8 x + 12$$

Solve for x

$$x=-2$$

$$x=2$$

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

Subtract 8x from both sides.

$$2x^{3}+3x^{2}-8x=12$$

Subtract 12 from both sides.

$$2x^{3}+3x^{2}-8x-12=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -12 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±6,±12,±3,±2,±4,±\frac{3}{2},±1,±\frac{1}{2}$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}+3x^{2}-8x-12 by x-2 to get 2x^{2}+7x+6. Solve the equation where the result equals to 0.

$$2x^{2}+7x+6=0$$

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, 7 for b, and 6 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 6}}{2\times 2}$$

Do the calculations.

$$x=\frac{-7±1}{4}$$

Solve the equation 2x^{2}+7x+6=0 when ± is plus and when ± is minus.

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

List all found solutions.

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