Solve x³ - x² - x + 1 = 0
The solutions of x³ - x² - x + 1 = 0 are x = -1, 1, 1 (repeated roots counted once each: -1, 1).
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 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
What is left is the quadratic x² - 1 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - x^{2} - x + 1 = (x + 1)(x - 1)^{2} = 0 \]
\[ x^{3} - x^{2} - x + 1 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & -1 & -1 & 1 \\ & 0 & 1 & 0 & -1 \\ \hline & 1 & 0 & -1 & 0 \end{array} \]
\[ x^{2} - 1 = 0 \Rightarrow x = 1,\ -1 \]
\[ x^{3} - x^{2} - x + 1 = (x + 1)(x - 1)^{2} \]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 | 1 | 1 |
| Product of the roots | -1 | -1 |
Questions
What are the roots of x³ - x² - x + 1 = 0?
The solutions of x³ - x² - x + 1 = 0 are x = -1, 1, 1 (repeated roots counted once each: -1, 1).
How do you factor x³ - x² - x + 1?
(x + 1)(x − 1).
What is the sum of the roots?
1 (it equals −(-1) by Vieta's formulas).