Solve x⁴ - 5x³ - 18x² + 128x - 160 = 0
The solutions of x⁴ - 5x³ - 18x² + 128x - 160 = 0 are x = -5, 2, 4, 4 (repeated roots counted once each: -5, 2, 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 -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
\[ x^{4} - 5x^{3} - 18x^{2} + 128x - 160 = (x + 5)(x - 2)(x - 4)^{2} = 0 \]
\[ x^{4} - 5x^{3} - 18x^{2} + 128x - 160 = 0 \]
\[ \begin{array}{c|ccccc} 2 & 1 & -5 & -18 & 128 & -160 \\ & 0 & 2 & -6 & -48 & 160 \\ \hline & 1 & -3 & -24 & 80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & -3 & -24 & 80 \\ & 0 & 4 & 4 & -80 \\ \hline & 1 & 1 & -20 & 0 \end{array} \]
\[ x^{2} + x - 20 = 0 \Rightarrow x = 4,\ -5 \]
\[ x^{4} - 5x^{3} - 18x^{2} + 128x - 160 = (x + 5)(x - 2)(x - 4)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -54 | no |
| -1 | -300 | 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 = -5, 2, 4, 4 (repeated roots counted once each: -5, 2, 4).
How do you factor x⁴ - 5x³ - 18x² + 128x - 160?
(x + 5)(x − 2)(x − 4).
What is the sum of the roots?
5 (it equals −(-5) by Vieta's formulas).