Solve x⁴ + x³ - 36x² - 16x + 320 = 0
The solutions of x⁴ + x³ - 36x² - 16x + 320 = 0 are x = -5, -4, 4, 4 (repeated roots counted once each: -5, -4, 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 320, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±32, ±40, ±64, ±80, ±160, ±320.
- 2
Test the candidates until one gives 0. x = 4 is a root. Divide by (x − 4) 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
\[ x^{4} + x^{3} - 36x^{2} - 16x + 320 = (x + 5)(x + 4)(x - 4)^{2} = 0 \]
\[ x^{4} + x^{3} - 36x^{2} - 16x + 320 = 0 \]
\[ \begin{array}{c|ccccc} 4 & 1 & 1 & -36 & -16 & 320 \\ & 0 & 4 & 20 & -64 & -320 \\ \hline & 1 & 5 & -16 & -80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 4 & 1 & 5 & -16 & -80 \\ & 0 & 4 & 36 & 80 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{4} + x^{3} - 36x^{2} - 16x + 320 = (x + 5)(x + 4)(x - 4)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 270 | no |
| -1 | 300 | no |
| 2 | 168 | no |
| -2 | 216 | no |
| 4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -1 | -1 |
| Product of the roots | 320 | 320 |
Questions
What are the roots of x⁴ + x³ - 36x² - 16x + 320 = 0?
The solutions of x⁴ + x³ - 36x² - 16x + 320 = 0 are x = -5, -4, 4, 4 (repeated roots counted once each: -5, -4, 4).
How do you factor x⁴ + x³ - 36x² - 16x + 320?
(x + 5)(x + 4)(x − 4).
What is the sum of the roots?
-1 (it equals −1 by Vieta's formulas).