Solve x⁴ - 20x³ + 150x² - 500x + 625 = 0
The solutions of x⁴ - 20x³ + 150x² - 500x + 625 = 0 are x = 5, 5, 5, 5 (repeated roots counted once each: 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 625, so the candidates are ±1, ±5, ±25, ±125, ±625.
- 2
Test the candidates until one gives 0. x = 5 is a root. Divide by (x − 5) 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
\[ x^{4} - 20x^{3} + 150x^{2} - 500x + 625 = (x - 5)^{4} = 0 \]
\[ x^{4} - 20x^{3} + 150x^{2} - 500x + 625 = 0 \]
\[ \begin{array}{c|ccccc} 5 & 1 & -20 & 150 & -500 & 625 \\ & 0 & 5 & -75 & 375 & -625 \\ \hline & 1 & -15 & 75 & -125 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 5 & 1 & -15 & 75 & -125 \\ & 0 & 5 & -50 & 125 \\ \hline & 1 & -10 & 25 & 0 \end{array} \]
\[ x^{2} - 10x + 25 = 0 \Rightarrow x = 5 \]
\[ x^{4} - 20x^{3} + 150x^{2} - 500x + 625 = (x - 5)^{4} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 256 | no |
| -1 | 1296 | no |
| 5 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 20 | 20 |
| Product of the roots | 625 | 625 |
Questions
What are the roots of x⁴ - 20x³ + 150x² - 500x + 625 = 0?
The solutions of x⁴ - 20x³ + 150x² - 500x + 625 = 0 are x = 5, 5, 5, 5 (repeated roots counted once each: 5).
How do you factor x⁴ - 20x³ + 150x² - 500x + 625?
(x − 5).
What is the sum of the roots?
20 (it equals −(-20) by Vieta's formulas).