Solve 2x^3=5x^2+12x-27

$$2 x ^ { 3 } = 5 x ^ { 2 } + 12 x - 27$$

Solve for x

$$x=\frac{-\sqrt{73}-1}{4}\approx -2.386000936$$

$$x=3$$

$$x = \frac{\sqrt{73} - 1}{4} \approx 1.886000936$$

Solution Steps

Subtract 5x^{2} from both sides.

$$2x^{3}-5x^{2}=12x-27$$

Subtract 12x from both sides.

$$2x^{3}-5x^{2}-12x=-27$$

Add 27 to both sides.

$$2x^{3}-5x^{2}-12x+27=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 27 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±\frac{27}{2},±27,±\frac{9}{2},±9,±\frac{3}{2},±3,±\frac{1}{2},±1$$

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}-5x^{2}-12x+27 by x-3 to get 2x^{2}+x-9. Solve the equation where the result equals to 0.

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

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 -9 for c in the quadratic formula.

$$x=\frac{-1±\sqrt{1^{2}-4\times 2\left(-9\right)}}{2\times 2}$$

Do the calculations.

$$x=\frac{-1±\sqrt{73}}{4}$$

Solve the equation 2x^{2}+x-9=0 when ± is plus and when ± is minus.

$$x=\frac{-\sqrt{73}-1}{4} x=\frac{\sqrt{73}-1}{4}$$

List all found solutions.

$$x=3 x=\frac{-\sqrt{73}-1}{4} x=\frac{\sqrt{73}-1}{4}$$