Solve x⁴ + 15x³ + 82x² + 192x + 160 = 0
The solutions of x⁴ + 15x³ + 82x² + 192x + 160 = 0 are x = -5, -4, -4, -2 (repeated roots counted once each: -5, -4, -2).
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
\[ x^{4} + 15x^{3} + 82x^{2} + 192x + 160 = (x + 5)(x + 4)^{2}(x + 2) = 0 \]
\[ x^{4} + 15x^{3} + 82x^{2} + 192x + 160 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & 15 & 82 & 192 & 160 \\ & 0 & -2 & -26 & -112 & -160 \\ \hline & 1 & 13 & 56 & 80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -4 & 1 & 13 & 56 & 80 \\ & 0 & -4 & -36 & -80 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{4} + 15x^{3} + 82x^{2} + 192x + 160 = (x + 5)(x + 4)^{2}(x + 2) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 450 | no |
| -1 | 36 | no |
| 2 | 1008 | 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 = -5, -4, -4, -2 (repeated roots counted once each: -5, -4, -2).
How do you factor x⁴ + 15x³ + 82x² + 192x + 160?
(x + 5)(x + 4)(x + 2).
What is the sum of the roots?
-15 (it equals −15 by Vieta's formulas).