Solve x⁴ + x³ - 21x² - 9x + 108 = 0
The solutions of x⁴ + x³ - 21x² - 9x + 108 = 0 are x = -4, -3, 3, 3 (repeated roots counted once each: -4, -3, 3).
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 108, so the candidates are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±27, ±36, ±54, ±108.
- 2
Test the candidates until one gives 0. x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 7x + 12 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + x^{3} - 21x^{2} - 9x + 108 = (x + 4)(x + 3)(x - 3)^{2} = 0 \]
\[ x^{4} + x^{3} - 21x^{2} - 9x + 108 = 0 \]
\[ \begin{array}{c|ccccc} 3 & 1 & 1 & -21 & -9 & 108 \\ & 0 & 3 & 12 & -27 & -108 \\ \hline & 1 & 4 & -9 & -36 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 3 & 1 & 4 & -9 & -36 \\ & 0 & 3 & 21 & 36 \\ \hline & 1 & 7 & 12 & 0 \end{array} \]
\[ x^{2} + 7x + 12 = 0 \Rightarrow x = -3,\ -4 \]
\[ x^{4} + x^{3} - 21x^{2} - 9x + 108 = (x + 4)(x + 3)(x - 3)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 80 | no |
| -1 | 96 | no |
| 2 | 30 | no |
| -2 | 50 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -1 | -1 |
| Product of the roots | 108 | 108 |
Questions
What are the roots of x⁴ + x³ - 21x² - 9x + 108 = 0?
The solutions of x⁴ + x³ - 21x² - 9x + 108 = 0 are x = -4, -3, 3, 3 (repeated roots counted once each: -4, -3, 3).
How do you factor x⁴ + x³ - 21x² - 9x + 108?
(x + 4)(x + 3)(x − 3).
What is the sum of the roots?
-1 (it equals −1 by Vieta's formulas).