Solve x⁴ + 4x³ + 6x² + 4x + 1 = 0
The solutions of x⁴ + 4x³ + 6x² + 4x + 1 = 0 are x = -1, -1, -1, -1 (repeated roots counted once each: -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 1, so the candidates are ±1.
- 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 + 2x + 1 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 4x^{3} + 6x^{2} + 4x + 1 = (x + 1)^{4} = 0 \]
\[ x^{4} + 4x^{3} + 6x^{2} + 4x + 1 = 0 \]
\[ \begin{array}{c|ccccc} -1 & 1 & 4 & 6 & 4 & 1 \\ & 0 & -1 & -3 & -3 & -1 \\ \hline & 1 & 3 & 3 & 1 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -1 & 1 & 3 & 3 & 1 \\ & 0 & -1 & -2 & -1 \\ \hline & 1 & 2 & 1 & 0 \end{array} \]
\[ x^{2} + 2x + 1 = 0 \Rightarrow x = -1 \]
\[ x^{4} + 4x^{3} + 6x^{2} + 4x + 1 = (x + 1)^{4} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 16 | no |
| -1 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -4 | -4 |
| Product of the roots | 1 | 1 |
Questions
What are the roots of x⁴ + 4x³ + 6x² + 4x + 1 = 0?
The solutions of x⁴ + 4x³ + 6x² + 4x + 1 = 0 are x = -1, -1, -1, -1 (repeated roots counted once each: -1).
How do you factor x⁴ + 4x³ + 6x² + 4x + 1?
(x + 1).
What is the sum of the roots?
-4 (it equals −4 by Vieta's formulas).