Solve x⁴ - x³ - 36x² + 16x + 320 = 0
The solutions of x⁴ - x³ - 36x² + 16x + 320 = 0 are x = -4, -4, 4, 5 (repeated roots counted once each: -4, 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 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 - x - 20 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - x^{3} - 36x^{2} + 16x + 320 = (x + 4)^{2}(x - 4)(x - 5) = 0 \]
\[ x^{4} - x^{3} - 36x^{2} + 16x + 320 = 0 \]
\[ \begin{array}{c|ccccc} 4 & 1 & -1 & -36 & 16 & 320 \\ & 0 & 4 & 12 & -96 & -320 \\ \hline & 1 & 3 & -24 & -80 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -4 & 1 & 3 & -24 & -80 \\ & 0 & -4 & 4 & 80 \\ \hline & 1 & -1 & -20 & 0 \end{array} \]
\[ x^{2} - x - 20 = 0 \Rightarrow x = 5,\ -4 \]
\[ x^{4} - x^{3} - 36x^{2} + 16x + 320 = (x + 4)^{2}(x - 4)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 300 | no |
| -1 | 270 | no |
| 2 | 216 | no |
| -2 | 168 | 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 = -4, -4, 4, 5 (repeated roots counted once each: -4, 4, 5).
How do you factor x⁴ - x³ - 36x² + 16x + 320?
(x + 4)(x − 4)(x − 5).
What is the sum of the roots?
1 (it equals −(-1) by Vieta's formulas).