Solve 9x^3-3x^2-8x+4=0

$$9 { x }^{ 3 } -3 { x }^{ 2 } -8x+4 = 0$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-1$$

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 4 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±\frac{4}{9},±\frac{4}{3},±4,±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 9x^{3}-3x^{2}-8x+4 by x+1 to get 9x^{2}-12x+4. Solve the equation where the result equals to 0.

$$9x^{2}-12x+4=0$$

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 9 for a, -12 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\times 4}}{2\times 9}$$

Do the calculations.

$$x=\frac{12±0}{18}$$

Solutions are the same.

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

List all found solutions.

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