Solve x³ + 6x² + 5x - 12 = 0
The solutions of x³ + 6x² + 5x - 12 = 0 are x = -4, -3, 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 -12, so the candidates are ±1, ±2, ±3, ±4, ±6, ±12.
- 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 + 7x + 12 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 6x^{2} + 5x - 12 = (x + 4)(x + 3)(x - 1) = 0 \]
\[ x^{3} + 6x^{2} + 5x - 12 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & 6 & 5 & -12 \\ & 0 & 1 & 7 & 12 \\ \hline & 1 & 7 & 12 & 0 \end{array} \]
\[ x^{2} + 7x + 12 = 0 \Rightarrow x = -3,\ -4 \]
\[ x^{3} + 6x^{2} + 5x - 12 = (x + 4)(x + 3)(x - 1) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -6 | -6 |
| Product of the roots | 12 | 12 |
Questions
What are the roots of x³ + 6x² + 5x - 12 = 0?
The solutions of x³ + 6x² + 5x - 12 = 0 are x = -4, -3, 1.
How do you factor x³ + 6x² + 5x - 12?
(x + 4)(x + 3)(x − 1).
What is the sum of the roots?
-6 (it equals −6 by Vieta's formulas).