Solve x³ + 13x² + 48x + 36 = 0
The solutions of x³ + 13x² + 48x + 36 = 0 are x = -6, -6, -1 (repeated roots counted once each: -6, -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 36, so the candidates are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
- 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 + 12x + 36 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 13x^{2} + 48x + 36 = (x + 6)^{2}(x + 1) = 0 \]
\[ x^{3} + 13x^{2} + 48x + 36 = 0 \]
\[ \begin{array}{c|cccc} -1 & 1 & 13 & 48 & 36 \\ & 0 & -1 & -12 & -36 \\ \hline & 1 & 12 & 36 & 0 \end{array} \]
\[ x^{2} + 12x + 36 = 0 \Rightarrow x = -6 \]
\[ x^{3} + 13x^{2} + 48x + 36 = (x + 6)^{2}(x + 1) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 98 | no |
| -1 | 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 | -36 | -36 |
Questions
What are the roots of x³ + 13x² + 48x + 36 = 0?
The solutions of x³ + 13x² + 48x + 36 = 0 are x = -6, -6, -1 (repeated roots counted once each: -6, -1).
How do you factor x³ + 13x² + 48x + 36?
(x + 6)(x + 1).
What is the sum of the roots?
-13 (it equals −13 by Vieta's formulas).