Solve x⁴ + 6x³ + 4x² - 6x - 5 = 0
The solutions of x⁴ + 6x³ + 4x² - 6x - 5 = 0 are x = -5, -1, -1, 1 (repeated roots counted once each: -5, -1, 1).
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 -5, so the candidates are ±1, ±5.
- 2
Test the candidates until one gives 0. x = 1 is a root. Divide by (x − 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = -1 is a root. Divide by (x + 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 6x + 5 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 6x^{3} + 4x^{2} - 6x - 5 = (x + 5)(x + 1)^{2}(x - 1) = 0 \]
\[ x^{4} + 6x^{3} + 4x^{2} - 6x - 5 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 6 & 4 & -6 & -5 \\ & 0 & 1 & 7 & 11 & 5 \\ \hline & 1 & 7 & 11 & 5 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -1 & 1 & 7 & 11 & 5 \\ & 0 & -1 & -6 & -5 \\ \hline & 1 & 6 & 5 & 0 \end{array} \]
\[ x^{2} + 6x + 5 = 0 \Rightarrow x = -1,\ -5 \]
\[ x^{4} + 6x^{3} + 4x^{2} - 6x - 5 = (x + 5)(x + 1)^{2}(x - 1) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 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 | -5 | -5 |
Questions
What are the roots of x⁴ + 6x³ + 4x² - 6x - 5 = 0?
The solutions of x⁴ + 6x³ + 4x² - 6x - 5 = 0 are x = -5, -1, -1, 1 (repeated roots counted once each: -5, -1, 1).
How do you factor x⁴ + 6x³ + 4x² - 6x - 5?
(x + 5)(x + 1)(x − 1).
What is the sum of the roots?
-6 (it equals −6 by Vieta's formulas).