Solve x³ - 11x² + 39x - 45 = 0
The solutions of x³ - 11x² + 39x - 45 = 0 are x = 3, 3, 5 (repeated roots counted once each: 3, 5).
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
\[ x^{3} - 11x^{2} + 39x - 45 = (x - 3)^{2}(x - 5) = 0 \]
\[ x^{3} - 11x^{2} + 39x - 45 = 0 \]
\[ \begin{array}{c|cccc} 3 & 1 & -11 & 39 & -45 \\ & 0 & 3 & -24 & 45 \\ \hline & 1 & -8 & 15 & 0 \end{array} \]
\[ x^{2} - 8x + 15 = 0 \Rightarrow x = 5,\ 3 \]
\[ x^{3} - 11x^{2} + 39x - 45 = (x - 3)^{2}(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -16 | no |
| -1 | -96 | 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 = 3, 3, 5 (repeated roots counted once each: 3, 5).
How do you factor x³ - 11x² + 39x - 45?
(x − 3)(x − 5).
What is the sum of the roots?
11 (it equals −(-11) by Vieta's formulas).