Solve 9x^3+36x^2-4x-16=0

$$9 x ^ { 3 } + 36 x ^ { 2 } - 4 x - 16 = 0$$

Solve for x

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

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

$$x=-4$$

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 -16 and q divides the leading coefficient 9. List all candidates \frac{p}{q}.

$$±\frac{16}{9},±\frac{16}{3},±16,±\frac{8}{9},±\frac{8}{3},±8,±\frac{4}{9},±\frac{4}{3},±4,±\frac{2}{9},±\frac{2}{3},±2,±\frac{1}{9},±\frac{1}{3},±1$$

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.

$$x=-4$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 9x^{3}+36x^{2}-4x-16 by x+4 to get 9x^{2}-4. Solve the equation where the result equals to 0.

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

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, 0 for b, and -4 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

Do the calculations.

$$x=\frac{0±12}{18}$$

Solve the equation 9x^{2}-4=0 when ± is plus and when ± is minus.

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

List all found solutions.

$$x=-4 x=-\frac{2}{3} x=\frac{2}{3}$$