Solve x⁴ - 15x³ + 82x² - 192x + 160 = 0
The solutions of x⁴ - 15x³ + 82x² - 192x + 160 = 0 are x = 2, 4, 4, 5 (repeated roots counted once each: 2, 4, 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 - 9x + 20 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 15x^{3} + 82x^{2} - 192x + 160 = (x - 2)(x - 4)^{2}(x - 5) = 0 \]
\[ x^{4} - 15x^{3} + 82x^{2} - 192x + 160 = 0 \]
\[ \begin{array}{c|ccccc} 2 & 1 & -15 & 82 & -192 & 160 \\ & 0 & 2 & -26 & 112 & -160 \\ \hline & 1 & -13 & 56 & -80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & -13 & 56 & -80 \\ & 0 & 4 & -36 & 80 \\ \hline & 1 & -9 & 20 & 0 \end{array} \]
\[ x^{2} - 9x + 20 = 0 \Rightarrow x = 5,\ 4 \]
\[ x^{4} - 15x^{3} + 82x^{2} - 192x + 160 = (x - 2)(x - 4)^{2}(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 36 | no |
| -1 | 450 | no |
| 2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 15 | 15 |
| Product of the roots | 160 | 160 |
Questions
What are the roots of x⁴ - 15x³ + 82x² - 192x + 160 = 0?
The solutions of x⁴ - 15x³ + 82x² - 192x + 160 = 0 are x = 2, 4, 4, 5 (repeated roots counted once each: 2, 4, 5).
How do you factor x⁴ - 15x³ + 82x² - 192x + 160?
(x − 2)(x − 4)(x − 5).
What is the sum of the roots?
15 (it equals −(-15) by Vieta's formulas).