Solve 2x^3-27x^2+90x-88=0

$$2 x ^ { 3 } - 27 x ^ { 2 } + 90 x - 88 = 0$$

Solve for x

$$x = \frac{23 - \sqrt{177}}{4} \approx 2.423966326$$

$$x=2$$

$$x = \frac{\sqrt{177} + 23}{4} \approx 9.076033674$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -88 and q divides the leading coefficient 2. List all candidates \frac{p}{q}.

$$±44,±88,±22,±11,±\frac{11}{2},±4,±8,±2,±1,±\frac{1}{2}$$

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}-27x^{2}+90x-88 by x-2 to get 2x^{2}-23x+44. Solve the equation where the result equals to 0.

$$2x^{2}-23x+44=0$$

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, -23 for b, and 44 for c in the quadratic formula.

$$x=\frac{-\left(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 2\times 44}}{2\times 2}$$

Do the calculations.

$$x=\frac{23±\sqrt{177}}{4}$$

Solve the equation 2x^{2}-23x+44=0 when ± is plus and when ± is minus.

$$x=\frac{23-\sqrt{177}}{4} x=\frac{\sqrt{177}+23}{4}$$

List all found solutions.

$$x=2 x=\frac{23-\sqrt{177}}{4} x=\frac{\sqrt{177}+23}{4}$$