Solve x³ + 11x² + 39x + 45 = 0
The solutions of x³ + 11x² + 39x + 45 = 0 are x = -5, -3, -3 (repeated roots counted once each: -5, -3).
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 45, so the candidates are ±1, ±3, ±5, ±9, ±15, ±45.
- 2
Test the candidates until one gives 0. x = -3 is a root. Divide by (x + 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 8x + 15 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 11x^{2} + 39x + 45 = (x + 5)(x + 3)^{2} = 0 \]
\[ x^{3} + 11x^{2} + 39x + 45 = 0 \]
\[ \begin{array}{c|cccc} -3 & 1 & 11 & 39 & 45 \\ & 0 & -3 & -24 & -45 \\ \hline & 1 & 8 & 15 & 0 \end{array} \]
\[ x^{2} + 8x + 15 = 0 \Rightarrow x = -3,\ -5 \]
\[ x^{3} + 11x^{2} + 39x + 45 = (x + 5)(x + 3)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 96 | no |
| -1 | 16 | no |
| 3 | 288 | no |
| -3 | 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 | -45 | -45 |
Questions
What are the roots of x³ + 11x² + 39x + 45 = 0?
The solutions of x³ + 11x² + 39x + 45 = 0 are x = -5, -3, -3 (repeated roots counted once each: -5, -3).
How do you factor x³ + 11x² + 39x + 45?
(x + 5)(x + 3).
What is the sum of the roots?
-11 (it equals −11 by Vieta's formulas).