Solve x⁴ + 12x³ + 54x² + 108x + 81 = 0
The solutions of x⁴ + 12x³ + 54x² + 108x + 81 = 0 are x = -3, -3, -3, -3 (repeated roots counted once each: -3).
How to work it out
- 1
A quartic has up to 4 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 81, so the candidates are ±1, ±3, ±9, ±27, ±81.
- 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
x = -3 is a root. Divide by (x + 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 6x + 9 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 12x^{3} + 54x^{2} + 108x + 81 = (x + 3)^{4} = 0 \]
\[ x^{4} + 12x^{3} + 54x^{2} + 108x + 81 = 0 \]
\[ \begin{array}{c|ccccc} -3 & 1 & 12 & 54 & 108 & 81 \\ & 0 & -3 & -27 & -81 & -81 \\ \hline & 1 & 9 & 27 & 27 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 9 & 27 & 27 \\ & 0 & -3 & -18 & -27 \\ \hline & 1 & 6 & 9 & 0 \end{array} \]
\[ x^{2} + 6x + 9 = 0 \Rightarrow x = -3 \]
\[ x^{4} + 12x^{3} + 54x^{2} + 108x + 81 = (x + 3)^{4} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 256 | no |
| -1 | 16 | no |
| 3 | 1296 | no |
| -3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -12 | -12 |
| Product of the roots | 81 | 81 |
Questions
What are the roots of x⁴ + 12x³ + 54x² + 108x + 81 = 0?
The solutions of x⁴ + 12x³ + 54x² + 108x + 81 = 0 are x = -3, -3, -3, -3 (repeated roots counted once each: -3).
How do you factor x⁴ + 12x³ + 54x² + 108x + 81?
(x + 3).
What is the sum of the roots?
-12 (it equals −12 by Vieta's formulas).