Solve x⁴ + 13x³ + 51x² + 35x - 100 = 0
The solutions of x⁴ + 13x³ + 51x² + 35x - 100 = 0 are x = -5, -5, -4, 1 (repeated roots counted once each: -5, -4, 1).
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
\[ x^{4} + 13x^{3} + 51x^{2} + 35x - 100 = (x + 5)^{2}(x + 4)(x - 1) = 0 \]
\[ x^{4} + 13x^{3} + 51x^{2} + 35x - 100 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 13 & 51 & 35 & -100 \\ & 0 & 1 & 14 & 65 & 100 \\ \hline & 1 & 14 & 65 & 100 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -4 & 1 & 14 & 65 & 100 \\ & 0 & -4 & -40 & -100 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{4} + 13x^{3} + 51x^{2} + 35x - 100 = (x + 5)^{2}(x + 4)(x - 1) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 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 = -5, -5, -4, 1 (repeated roots counted once each: -5, -4, 1).
How do you factor x⁴ + 13x³ + 51x² + 35x - 100?
(x + 5)(x + 4)(x − 1).
What is the sum of the roots?
-13 (it equals −13 by Vieta's formulas).