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 | 1296 | no |
| -1 | 256 | no |
| 5 | 10000 | 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).