Solve a^3-11a^2-9a+99=0

$$a ^ { 3 } - 11 a ^ { 2 } - 9 a + 99 = 0$$

Solve for a

$$a=3$$

$$a=11$$

$$a=-3$$

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 99 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±99,±33,±11,±9,±3,±1$$

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.

$$a=3$$

By Factor theorem, a-k is a factor of the polynomial for each root k. Divide a^{3}-11a^{2}-9a+99 by a-3 to get a^{2}-8a-33. Solve the equation where the result equals to 0.

$$a^{2}-8a-33=0$$

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 1 for a, -8 for b, and -33 for c in the quadratic formula.

$$a=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 1\left(-33\right)}}{2}$$

Do the calculations.

$$a=\frac{8±14}{2}$$

Solve the equation a^{2}-8a-33=0 when ± is plus and when ± is minus.

$$a=-3 a=11$$

List all found solutions.

$$a=3 a=-3 a=11$$