Solve x³ + 7x² + 11x + 5 = 0
The solutions of x³ + 7x² + 11x + 5 = 0 are x = -5, -1, -1 (repeated roots counted once each: -5, -1).
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 5, so the candidates are ±1, ±5.
- 2
Test the candidates until one gives 0. x = -1 is a root. Divide by (x + 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 6x + 5 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 7x^{2} + 11x + 5 = (x + 5)(x + 1)^{2} = 0 \]
\[ x^{3} + 7x^{2} + 11x + 5 = 0 \]
\[ \begin{array}{c|cccc} -1 & 1 & 7 & 11 & 5 \\ & 0 & -1 & -6 & -5 \\ \hline & 1 & 6 & 5 & 0 \end{array} \]
\[ x^{2} + 6x + 5 = 0 \Rightarrow x = -1,\ -5 \]
\[ x^{3} + 7x^{2} + 11x + 5 = (x + 5)(x + 1)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 24 | no |
| -1 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -7 | -7 |
| Product of the roots | -5 | -5 |
Questions
What are the roots of x³ + 7x² + 11x + 5 = 0?
The solutions of x³ + 7x² + 11x + 5 = 0 are x = -5, -1, -1 (repeated roots counted once each: -5, -1).
How do you factor x³ + 7x² + 11x + 5?
(x + 5)(x + 1).
What is the sum of the roots?
-7 (it equals −7 by Vieta's formulas).