Solve x⁴ - 11x³ + 15x² + 175x - 500 = 0
The solutions of x⁴ - 11x³ + 15x² + 175x - 500 = 0 are x = -4, 5, 5, 5 (repeated roots counted once each: -4, 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 -500, so the candidates are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, ±100, ±125, ±250, ±500.
- 2
Test the candidates until one gives 0. x = -4 is a root. Divide by (x + 4) 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
\[ x^{4} - 11x^{3} + 15x^{2} + 175x - 500 = (x + 4)(x - 5)^{3} = 0 \]
\[ x^{4} - 11x^{3} + 15x^{2} + 175x - 500 = 0 \]
\[ \begin{array}{c|ccccc} -4 & 1 & -11 & 15 & 175 & -500 \\ & 0 & -4 & 60 & -300 & 500 \\ \hline & 1 & -15 & 75 & -125 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 5 & 1 & -15 & 75 & -125 \\ & 0 & 5 & -50 & 125 \\ \hline & 1 & -10 & 25 & 0 \end{array} \]
\[ x^{2} - 10x + 25 = 0 \Rightarrow x = 5 \]
\[ x^{4} - 11x^{3} + 15x^{2} + 175x - 500 = (x + 4)(x - 5)^{3} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -320 | no |
| -1 | -648 | no |
| 2 | -162 | no |
| -2 | -686 | no |
| 4 | -8 | no |
| -4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 11 | 11 |
| Product of the roots | -500 | -500 |
Questions
What are the roots of x⁴ - 11x³ + 15x² + 175x - 500 = 0?
The solutions of x⁴ - 11x³ + 15x² + 175x - 500 = 0 are x = -4, 5, 5, 5 (repeated roots counted once each: -4, 5).
How do you factor x⁴ - 11x³ + 15x² + 175x - 500?
(x + 4)(x − 5).
What is the sum of the roots?
11 (it equals −(-11) by Vieta's formulas).