Solve 2x^3+x^2-18x=9

$$2 x ^ { 3 } + x ^ { 2 } - 18 x = 9$$

Solve for x

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

$$x=3$$

$$x=-3$$

Solution Steps

Subtract 9 from both sides.

$$2x^{3}+x^{2}-18x-9=0$$

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}+x^{2}-18x-9 by x-3 to get 2x^{2}+7x+3. Solve the equation where the result equals to 0.

$$2x^{2}+7x+3=0$$

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

$$x=\frac{-7±\sqrt{7^{2}-4\times 2\times 3}}{2\times 2}$$

Do the calculations.

$$x=\frac{-7±5}{4}$$

Solve the equation 2x^{2}+7x+3=0 when ± is plus and when ± is minus.

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

List all found solutions.

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