Solve 2x^3-3x^2-12x=-9

$$2 { x }^{ 3 } -3 { x }^{ 2 } -12x=-9$$

Solve for x

$$x=\frac{-\sqrt{33}-3}{4}\approx -2.186140662$$

$$x=3$$

$$x=\frac{\sqrt{33}-3}{4}\approx 0.686140662$$

Solution Steps

Add 9 to both sides.

$$2x^{3}-3x^{2}-12x+9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 9 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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.

$$x=3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 2x^{3}-3x^{2}-12x+9 by x-3 to get 2x^{2}+3x-3. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-3±\sqrt{33}}{4}$$

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

$$x=\frac{-\sqrt{33}-3}{4} x=\frac{\sqrt{33}-3}{4}$$

List all found solutions.

$$x=3 x=\frac{-\sqrt{33}-3}{4} x=\frac{\sqrt{33}-3}{4}$$