Solve x⁴ - 10x³ + 21x² + 40x - 100 = 0
The solutions of x⁴ - 10x³ + 21x² + 40x - 100 = 0 are x = -2, 2, 5, 5 (repeated roots counted once each: -2, 2, 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 = 2 is a root. Divide by (x − 2) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = -2 is a root. Divide by (x + 2) 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
\[ x^{4} - 10x^{3} + 21x^{2} + 40x - 100 = (x + 2)(x - 2)(x - 5)^{2} = 0 \]
\[ x^{4} - 10x^{3} + 21x^{2} + 40x - 100 = 0 \]
\[ \begin{array}{c|ccccc} 2 & 1 & -10 & 21 & 40 & -100 \\ & 0 & 2 & -16 & 10 & 100 \\ \hline & 1 & -8 & 5 & 50 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & -8 & 5 & 50 \\ & 0 & -2 & 20 & -50 \\ \hline & 1 & -10 & 25 & 0 \end{array} \]
\[ x^{2} - 10x + 25 = 0 \Rightarrow x = 5 \]
\[ x^{4} - 10x^{3} + 21x^{2} + 40x - 100 = (x + 2)(x - 2)(x - 5)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -48 | no |
| -1 | -108 | no |
| 2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 10 | 10 |
| Product of the roots | -100 | -100 |
Questions
What are the roots of x⁴ - 10x³ + 21x² + 40x - 100 = 0?
The solutions of x⁴ - 10x³ + 21x² + 40x - 100 = 0 are x = -2, 2, 5, 5 (repeated roots counted once each: -2, 2, 5).
How do you factor x⁴ - 10x³ + 21x² + 40x - 100?
(x + 2)(x − 2)(x − 5).
What is the sum of the roots?
10 (it equals −(-10) by Vieta's formulas).