Solve z^3-9z^2+z-9=0

$$z ^ { 3 } - 9 z ^ { 2 } + z - 9 = 0$$

Solve for z

$$z=9$$

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 -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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.

$$z=9$$

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

$$z^{2}+1=0$$

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, 0 for b, and 1 for c in the quadratic formula.

$$z=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$z=\frac{0±\sqrt{-4}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$z\in \emptyset$$

List all found solutions.

$$z=9$$

Solve for z (complex solution)

$$z=-i$$

$$z=9$$

$$z=i$$