Solve x⁴ + 2x³ - 12x² + 14x - 5 = 0
The solutions of x⁴ + 2x³ - 12x² + 14x - 5 = 0 are x = -5, 1, 1, 1 (repeated roots counted once each: -5, 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 + 4x - 5 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 2x^{3} - 12x^{2} + 14x - 5 = (x + 5)(x - 1)^{3} = 0 \]
\[ x^{4} + 2x^{3} - 12x^{2} + 14x - 5 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 2 & -12 & 14 & -5 \\ & 0 & 1 & 3 & -9 & 5 \\ \hline & 1 & 3 & -9 & 5 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 1 & 1 & 3 & -9 & 5 \\ & 0 & 1 & 4 & -5 \\ \hline & 1 & 4 & -5 & 0 \end{array} \]
\[ x^{2} + 4x - 5 = 0 \Rightarrow x = 1,\ -5 \]
\[ x^{4} + 2x^{3} - 12x^{2} + 14x - 5 = (x + 5)(x - 1)^{3} \]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 | -2 | -2 |
| Product of the roots | -5 | -5 |
Questions
What are the roots of x⁴ + 2x³ - 12x² + 14x - 5 = 0?
The solutions of x⁴ + 2x³ - 12x² + 14x - 5 = 0 are x = -5, 1, 1, 1 (repeated roots counted once each: -5, 1).
How do you factor x⁴ + 2x³ - 12x² + 14x - 5?
(x + 5)(x − 1).
What is the sum of the roots?
-2 (it equals −2 by Vieta's formulas).