Solve x³ + 11x² + 38x + 40 = 0
The solutions of x³ + 11x² + 38x + 40 = 0 are x = -5, -4, -2.
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 40, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40.
- 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
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
\[ x^{3} + 11x^{2} + 38x + 40 = (x + 5)(x + 4)(x + 2) = 0 \]
\[ x^{3} + 11x^{2} + 38x + 40 = 0 \]
\[ \begin{array}{c|cccc} -2 & 1 & 11 & 38 & 40 \\ & 0 & -2 & -18 & -40 \\ \hline & 1 & 9 & 20 & 0 \end{array} \]
\[ x^{2} + 9x + 20 = 0 \Rightarrow x = -4,\ -5 \]
\[ x^{3} + 11x^{2} + 38x + 40 = (x + 5)(x + 4)(x + 2) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 90 | no |
| -1 | 12 | no |
| 2 | 168 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -11 | -11 |
| Product of the roots | -40 | -40 |
Questions
What are the roots of x³ + 11x² + 38x + 40 = 0?
The solutions of x³ + 11x² + 38x + 40 = 0 are x = -5, -4, -2.
How do you factor x³ + 11x² + 38x + 40?
(x + 5)(x + 4)(x + 2).
What is the sum of the roots?
-11 (it equals −11 by Vieta's formulas).