Solve x⁴ - 13x³ + 51x² - 35x - 100 = 0
The solutions of x⁴ - 13x³ + 51x² - 35x - 100 = 0 are x = -1, 4, 5, 5 (repeated roots counted once each: -1, 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 -100, so the candidates are ±1, ±2, ±4, ±5, ±10, ±20, ±25, ±50, ±100.
- 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 = 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
\[ x^{4} - 13x^{3} + 51x^{2} - 35x - 100 = (x + 1)(x - 4)(x - 5)^{2} = 0 \]
\[ x^{4} - 13x^{3} + 51x^{2} - 35x - 100 = 0 \]
\[ \begin{array}{c|ccccc} -1 & 1 & -13 & 51 & -35 & -100 \\ & 0 & -1 & 14 & -65 & 100 \\ \hline & 1 & -14 & 65 & -100 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & -14 & 65 & -100 \\ & 0 & 4 & -40 & 100 \\ \hline & 1 & -10 & 25 & 0 \end{array} \]
\[ x^{2} - 10x + 25 = 0 \Rightarrow x = 5 \]
\[ x^{4} - 13x^{3} + 51x^{2} - 35x - 100 = (x + 1)(x - 4)(x - 5)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -96 | no |
| -1 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 13 | 13 |
| Product of the roots | -100 | -100 |
Questions
What are the roots of x⁴ - 13x³ + 51x² - 35x - 100 = 0?
The solutions of x⁴ - 13x³ + 51x² - 35x - 100 = 0 are x = -1, 4, 5, 5 (repeated roots counted once each: -1, 4, 5).
How do you factor x⁴ - 13x³ + 51x² - 35x - 100?
(x + 1)(x − 4)(x − 5).
What is the sum of the roots?
13 (it equals −(-13) by Vieta's formulas).