Solve x³ + 13x² + 56x + 80 = 0
The solutions of x³ + 13x² + 56x + 80 = 0 are x = -5, -4, -4 (repeated roots counted once each: -5, -4).
How to work it out
- 1
A cubic has up to 3 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 80, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±16, ±20, ±40, ±80.
- 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
What is left is the quadratic x² + 9x + 20 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 13x^{2} + 56x + 80 = (x + 5)(x + 4)^{2} = 0 \]
\[ x^{3} + 13x^{2} + 56x + 80 = 0 \]
\[ \begin{array}{c|cccc} -4 & 1 & 13 & 56 & 80 \\ & 0 & -4 & -36 & -80 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{3} + 13x^{2} + 56x + 80 = (x + 5)(x + 4)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 150 | no |
| -1 | 36 | no |
| 2 | 252 | no |
| -2 | 12 | no |
| 4 | 576 | no |
| -4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -13 | -13 |
| Product of the roots | -80 | -80 |
Questions
What are the roots of x³ + 13x² + 56x + 80 = 0?
The solutions of x³ + 13x² + 56x + 80 = 0 are x = -5, -4, -4 (repeated roots counted once each: -5, -4).
How do you factor x³ + 13x² + 56x + 80?
(x + 5)(x + 4).
What is the sum of the roots?
-13 (it equals −13 by Vieta's formulas).