Solve x³ + 10x² + 19x - 30 = 0
The solutions of x³ + 10x² + 19x - 30 = 0 are x = -6, -5, 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 -30, so the candidates are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±30.
- 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 + 11x + 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 10x^{2} + 19x - 30 = (x + 6)(x + 5)(x - 1) = 0 \]
\[ x^{3} + 10x^{2} + 19x - 30 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & 10 & 19 & -30 \\ & 0 & 1 & 11 & 30 \\ \hline & 1 & 11 & 30 & 0 \end{array} \]
\[ x^{2} + 11x + 30 = 0 \Rightarrow x = -5,\ -6 \]
\[ x^{3} + 10x^{2} + 19x - 30 = (x + 6)(x + 5)(x - 1) \]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 | -10 | -10 |
| Product of the roots | 30 | 30 |
Questions
What are the roots of x³ + 10x² + 19x - 30 = 0?
The solutions of x³ + 10x² + 19x - 30 = 0 are x = -6, -5, 1.
How do you factor x³ + 10x² + 19x - 30?
(x + 6)(x + 5)(x − 1).
What is the sum of the roots?
-10 (it equals −10 by Vieta's formulas).