Solve x⁴ - 18x³ + 121x² - 360x + 400 = 0
The solutions of x⁴ - 18x³ + 121x² - 360x + 400 = 0 are x = 4, 4, 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 400, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±25, ±40, ±50, ±80, ±100, ±200, ±400.
- 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 = 4 is a root. Divide by (x − 4) 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
\[ x^{4} - 18x^{3} + 121x^{2} - 360x + 400 = (x - 4)^{2}(x - 5)^{2} = 0 \]
\[ x^{4} - 18x^{3} + 121x^{2} - 360x + 400 = 0 \]
\[ \begin{array}{c|ccccc} 4 & 1 & -18 & 121 & -360 & 400 \\ & 0 & 4 & -56 & 260 & -400 \\ \hline & 1 & -14 & 65 & -100 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & -14 & 65 & -100 \\ & 0 & 4 & -40 & 100 \\ \hline & 1 & -10 & 25 & 0 \end{array} \]
\[ x^{2} - 10x + 25 = 0 \Rightarrow x = 5 \]
\[ x^{4} - 18x^{3} + 121x^{2} - 360x + 400 = (x - 4)^{2}(x - 5)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 144 | no |
| -1 | 900 | no |
| 2 | 36 | no |
| -2 | 1764 | no |
| 4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 18 | 18 |
| Product of the roots | 400 | 400 |
Questions
What are the roots of x⁴ - 18x³ + 121x² - 360x + 400 = 0?
The solutions of x⁴ - 18x³ + 121x² - 360x + 400 = 0 are x = 4, 4, 5, 5 (repeated roots counted once each: 4, 5).
How do you factor x⁴ - 18x³ + 121x² - 360x + 400?
(x − 4)(x − 5).
What is the sum of the roots?
18 (it equals −(-18) by Vieta's formulas).