Solve x³ + 10x² + 31x + 30 = 0
The solutions of x³ + 10x² + 31x + 30 = 0 are x = -5, -3, -2.
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 30, so the candidates are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30.
- 2
Test the candidates until one gives 0. x = -2 is a root. Divide by (x + 2) 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
\[ x^{3} + 10x^{2} + 31x + 30 = (x + 5)(x + 3)(x + 2) = 0 \]
\[ x^{3} + 10x^{2} + 31x + 30 = 0 \]
\[ \begin{array}{c|cccc} -2 & 1 & 10 & 31 & 30 \\ & 0 & -2 & -16 & -30 \\ \hline & 1 & 8 & 15 & 0 \end{array} \]
\[ x^{2} + 8x + 15 = 0 \Rightarrow x = -3,\ -5 \]
\[ x^{3} + 10x^{2} + 31x + 30 = (x + 5)(x + 3)(x + 2) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 72 | no |
| -1 | 8 | no |
| 2 | 140 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -10 | -10 |
| Product of the roots | -30 | -30 |
Questions
What are the roots of x³ + 10x² + 31x + 30 = 0?
The solutions of x³ + 10x² + 31x + 30 = 0 are x = -5, -3, -2.
How do you factor x³ + 10x² + 31x + 30?
(x + 5)(x + 3)(x + 2).
What is the sum of the roots?
-10 (it equals −10 by Vieta's formulas).