Solve x³ - 6x² - 36x + 216 = 0
The solutions of x³ - 6x² - 36x + 216 = 0 are x = -6, 6, 6 (repeated roots counted once each: -6, 6).
How to work it out
- 1
A cubic has up to 3 real roots. The rational root theorem says that any integer root of a polynomial with integer coefficients and leading coefficient 1 divides the constant term 216, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±27, ±36, ±54, ±72, ±108, ±216.
- 2
Test the candidates until one gives 0. x = 6 is a root. Divide by (x − 6) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² - 36 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - 6x^{2} - 36x + 216 = (x + 6)(x - 6)^{2} = 0 \]
\[ x^{3} - 6x^{2} - 36x + 216 = 0 \]
\[ \begin{array}{c|cccc} 6 & 1 & -6 & -36 & 216 \\ & 0 & 6 & 0 & -216 \\ \hline & 1 & 0 & -36 & 0 \end{array} \]
\[ x^{2} - 36 = 0 \Rightarrow x = 6,\ -6 \]
\[ x^{3} - 6x^{2} - 36x + 216 = (x + 6)(x - 6)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 175 | no |
| -1 | 245 | no |
| 2 | 128 | no |
| -2 | 256 | no |
| 3 | 81 | no |
| -3 | 243 | no |
| 4 | 40 | no |
| -4 | 200 | no |
| 6 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 6 | 6 |
| Product of the roots | -216 | -216 |
Questions
What are the roots of x³ - 6x² - 36x + 216 = 0?
The solutions of x³ - 6x² - 36x + 216 = 0 are x = -6, 6, 6 (repeated roots counted once each: -6, 6).
How do you factor x³ - 6x² - 36x + 216?
(x + 6)(x − 6).
What is the sum of the roots?
6 (it equals −(-6) by Vieta's formulas).