Solve x⁴ + 5x³ - 18x² - 128x - 160 = 0
The solutions of x⁴ + 5x³ - 18x² - 128x - 160 = 0 are x = -4, -4, -2, 5 (repeated roots counted once each: -4, -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 -160, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±32, ±40, ±80, ±160.
- 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 = -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 - x - 20 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 5x^{3} - 18x^{2} - 128x - 160 = (x + 4)^{2}(x + 2)(x - 5) = 0 \]
\[ x^{4} + 5x^{3} - 18x^{2} - 128x - 160 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & 5 & -18 & -128 & -160 \\ & 0 & -2 & -6 & 48 & 160 \\ \hline & 1 & 3 & -24 & -80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -4 & 1 & 3 & -24 & -80 \\ & 0 & -4 & 4 & 80 \\ \hline & 1 & -1 & -20 & 0 \end{array} \]
\[ x^{2} - x - 20 = 0 \Rightarrow x = 5,\ -4 \]
\[ x^{4} + 5x^{3} - 18x^{2} - 128x - 160 = (x + 4)^{2}(x + 2)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -300 | no |
| -1 | -54 | no |
| 2 | -432 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -5 | -5 |
| Product of the roots | -160 | -160 |
Questions
What are the roots of x⁴ + 5x³ - 18x² - 128x - 160 = 0?
The solutions of x⁴ + 5x³ - 18x² - 128x - 160 = 0 are x = -4, -4, -2, 5 (repeated roots counted once each: -4, -2, 5).
How do you factor x⁴ + 5x³ - 18x² - 128x - 160?
(x + 4)(x + 2)(x − 5).
What is the sum of the roots?
-5 (it equals −5 by Vieta's formulas).