Solve k^3-8k^2+21k-18=0

$$k ^ { 3 } - 8 k ^ { 2 } + 21 k - 18 = 0$$

Solve for k

$$k=3$$

$$k=2$$

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 -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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.

$$k=2$$

By Factor theorem, k-k is a factor of the polynomial for each root k. Divide k^{3}-8k^{2}+21k-18 by k-2 to get k^{2}-6k+9. Solve the equation where the result equals to 0.

$$k^{2}-6k+9=0$$

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, -6 for b, and 9 for c in the quadratic formula.

$$k=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$k=\frac{6±0}{2}$$

Solutions are the same.

$$k=3$$

List all found solutions.

$$k=2 k=3$$