Solve x³ + 8x² + 5x - 50 = 0
The solutions of x³ + 8x² + 5x - 50 = 0 are x = -5, -5, 2 (repeated roots counted once each: -5, 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 -50, so the candidates are ±1, ±2, ±5, ±10, ±25, ±50.
- 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 + 10x + 25 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 8x^{2} + 5x - 50 = (x + 5)^{2}(x - 2) = 0 \]
\[ x^{3} + 8x^{2} + 5x - 50 = 0 \]
\[ \begin{array}{c|cccc} 2 & 1 & 8 & 5 & -50 \\ & 0 & 2 & 20 & 50 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{3} + 8x^{2} + 5x - 50 = (x + 5)^{2}(x - 2) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -36 | no |
| -1 | -48 | no |
| 2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -8 | -8 |
| Product of the roots | 50 | 50 |
Questions
What are the roots of x³ + 8x² + 5x - 50 = 0?
The solutions of x³ + 8x² + 5x - 50 = 0 are x = -5, -5, 2 (repeated roots counted once each: -5, 2).
How do you factor x³ + 8x² + 5x - 50?
(x + 5)(x − 2).
What is the sum of the roots?
-8 (it equals −8 by Vieta's formulas).