Solve x³ - 27x + 54 = 0
The solutions of x³ - 27x + 54 = 0 are x = -6, 3, 3 (repeated roots counted once each: -6, 3).
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 54, so the candidates are ±1, ±2, ±3, ±6, ±9, ±18, ±27, ±54.
- 2
Test the candidates until one gives 0. x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 3x - 18 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - 27x + 54 = (x + 6)(x - 3)^{2} = 0 \]
\[ x^{3} - 27x + 54 = 0 \]
\[ \begin{array}{c|cccc} 3 & 1 & 0 & -27 & 54 \\ & 0 & 3 & 9 & -54 \\ \hline & 1 & 3 & -18 & 0 \end{array} \]
\[ x^{2} + 3x - 18 = 0 \Rightarrow x = 3,\ -6 \]
\[ x^{3} - 27x + 54 = (x + 6)(x - 3)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 28 | no |
| -1 | 80 | no |
| 2 | 8 | no |
| -2 | 100 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 0 | 0 |
| Product of the roots | -54 | -54 |
Questions
What are the roots of x³ - 27x + 54 = 0?
The solutions of x³ - 27x + 54 = 0 are x = -6, 3, 3 (repeated roots counted once each: -6, 3).
How do you factor x³ - 27x + 54?
(x + 6)(x − 3).
What is the sum of the roots?
0 (it equals −0 by Vieta's formulas).