Solve x⁴ + 8x³ - x² - 128x - 240 = 0
The solutions of x⁴ + 8x³ - x² - 128x - 240 = 0 are x = -5, -4, -3, 4.
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 -240, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±16, ±20, ±24, ±30, ±40, ±48, ±60, ±80, ±120, ±240.
- 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 = 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 + 9x + 20 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 8x^{3} - x^{2} - 128x - 240 = (x + 5)(x + 4)(x + 3)(x - 4) = 0 \]
\[ x^{4} + 8x^{3} - x^{2} - 128x - 240 = 0 \]
\[ \begin{array}{c|ccccc} -3 & 1 & 8 & -1 & -128 & -240 \\ & 0 & -3 & -15 & 48 & 240 \\ \hline & 1 & 5 & -16 & -80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & 5 & -16 & -80 \\ & 0 & 4 & 36 & 80 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{4} + 8x^{3} - x^{2} - 128x - 240 = (x + 5)(x + 4)(x + 3)(x - 4) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -360 | no |
| -1 | -120 | no |
| 2 | -420 | no |
| -2 | -36 | no |
| 3 | -336 | no |
| -3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -8 | -8 |
| Product of the roots | -240 | -240 |
Questions
What are the roots of x⁴ + 8x³ - x² - 128x - 240 = 0?
The solutions of x⁴ + 8x³ - x² - 128x - 240 = 0 are x = -5, -4, -3, 4.
How do you factor x⁴ + 8x³ - x² - 128x - 240?
(x + 5)(x + 4)(x + 3)(x − 4).
What is the sum of the roots?
-8 (it equals −8 by Vieta's formulas).