Solve x⁴ + 9x³ + 11x² - 81x - 180 = 0
The solutions of x⁴ + 9x³ + 11x² - 81x - 180 = 0 are x = -5, -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 -180, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±9, ±10, ±12, ±15, ±18, ±20, ±30, ±36, ±45, ±60, ±90, ±180.
- 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 + 9x + 20 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 9x^{3} + 11x^{2} - 81x - 180 = (x + 5)(x + 4)(x + 3)(x - 3) = 0 \]
\[ x^{4} + 9x^{3} + 11x^{2} - 81x - 180 = 0 \]
\[ \begin{array}{c|ccccc} 3 & 1 & 9 & 11 & -81 & -180 \\ & 0 & 3 & 36 & 141 & 180 \\ \hline & 1 & 12 & 47 & 60 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 12 & 47 & 60 \\ & 0 & -3 & -27 & -60 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{4} + 9x^{3} + 11x^{2} - 81x - 180 = (x + 5)(x + 4)(x + 3)(x - 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -240 | no |
| -1 | -96 | no |
| 2 | -210 | no |
| -2 | -30 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -9 | -9 |
| Product of the roots | -180 | -180 |
Questions
What are the roots of x⁴ + 9x³ + 11x² - 81x - 180 = 0?
The solutions of x⁴ + 9x³ + 11x² - 81x - 180 = 0 are x = -5, -4, -3, 3.
How do you factor x⁴ + 9x³ + 11x² - 81x - 180?
(x + 5)(x + 4)(x + 3)(x − 3).
What is the sum of the roots?
-9 (it equals −9 by Vieta's formulas).