Solve x³ + 4x² - 27x - 90 = 0
The solutions of x³ + 4x² - 27x - 90 = 0 are x = -6, -3, 5.
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 -90, so the candidates are ±1, ±2, ±3, ±5, ±6, ±9, ±10, ±15, ±18, ±30, ±45, ±90.
- 2
Test the candidates until one gives 0. x = -3 is a root. Divide by (x + 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + x - 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 4x^{2} - 27x - 90 = (x + 6)(x + 3)(x - 5) = 0 \]
\[ x^{3} + 4x^{2} - 27x - 90 = 0 \]
\[ \begin{array}{c|cccc} -3 & 1 & 4 & -27 & -90 \\ & 0 & -3 & -3 & 90 \\ \hline & 1 & 1 & -30 & 0 \end{array} \]
\[ x^{2} + x - 30 = 0 \Rightarrow x = 5,\ -6 \]
\[ x^{3} + 4x^{2} - 27x - 90 = (x + 6)(x + 3)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -112 | no |
| -1 | -60 | no |
| 2 | -120 | no |
| -2 | -28 | no |
| 3 | -108 | no |
| -3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -4 | -4 |
| Product of the roots | 90 | 90 |
Questions
What are the roots of x³ + 4x² - 27x - 90 = 0?
The solutions of x³ + 4x² - 27x - 90 = 0 are x = -6, -3, 5.
How do you factor x³ + 4x² - 27x - 90?
(x + 6)(x + 3)(x − 5).
What is the sum of the roots?
-4 (it equals −4 by Vieta's formulas).