Solve x⁴ + 6x³ - 20x² - 150x - 125 = 0
The solutions of x⁴ + 6x³ - 20x² - 150x - 125 = 0 are x = -5, -5, -1, 5 (repeated roots counted once each: -5, -1, 5).
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 -125, so the candidates are ±1, ±5, ±25, ±125.
- 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 = 5 is a root. Divide by (x − 5) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 10x + 25 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 6x^{3} - 20x^{2} - 150x - 125 = (x + 5)^{2}(x + 1)(x - 5) = 0 \]
\[ x^{4} + 6x^{3} - 20x^{2} - 150x - 125 = 0 \]
\[ \begin{array}{c|ccccc} -1 & 1 & 6 & -20 & -150 & -125 \\ & 0 & -1 & -5 & 25 & 125 \\ \hline & 1 & 5 & -25 & -125 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 5 & 1 & 5 & -25 & -125 \\ & 0 & 5 & 50 & 125 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{4} + 6x^{3} - 20x^{2} - 150x - 125 = (x + 5)^{2}(x + 1)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -288 | no |
| -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 | -125 | -125 |
Questions
What are the roots of x⁴ + 6x³ - 20x² - 150x - 125 = 0?
The solutions of x⁴ + 6x³ - 20x² - 150x - 125 = 0 are x = -5, -5, -1, 5 (repeated roots counted once each: -5, -1, 5).
How do you factor x⁴ + 6x³ - 20x² - 150x - 125?
(x + 5)(x + 1)(x − 5).
What is the sum of the roots?
-6 (it equals −6 by Vieta's formulas).