Solve x³ + 14x² + 63x + 90 = 0
The solutions of x³ + 14x² + 63x + 90 = 0 are x = -6, -5, -3.
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 + 11x + 30 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 14x^{2} + 63x + 90 = (x + 6)(x + 5)(x + 3) = 0 \]
\[ x^{3} + 14x^{2} + 63x + 90 = 0 \]
\[ \begin{array}{c|cccc} -3 & 1 & 14 & 63 & 90 \\ & 0 & -3 & -33 & -90 \\ \hline & 1 & 11 & 30 & 0 \end{array} \]
\[ x^{2} + 11x + 30 = 0 \Rightarrow x = -5,\ -6 \]
\[ x^{3} + 14x^{2} + 63x + 90 = (x + 6)(x + 5)(x + 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 168 | no |
| -1 | 40 | no |
| 2 | 280 | no |
| -2 | 12 | no |
| 3 | 432 | no |
| -3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -14 | -14 |
| Product of the roots | -90 | -90 |
Questions
What are the roots of x³ + 14x² + 63x + 90 = 0?
The solutions of x³ + 14x² + 63x + 90 = 0 are x = -6, -5, -3.
How do you factor x³ + 14x² + 63x + 90?
(x + 6)(x + 5)(x + 3).
What is the sum of the roots?
-14 (it equals −14 by Vieta's formulas).