Solve x⁴ + 7x³ + 13x² - 3x - 18 = 0
The solutions of x⁴ + 7x³ + 13x² - 3x - 18 = 0 are x = -3, -3, -2, 1 (repeated roots counted once each: -3, -2, 1).
How to work it out
- 1
A quartic has up to 4 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 -18, so the candidates are ±1, ±2, ±3, ±6, ±9, ±18.
- 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
x = -2 is a root. Divide by (x + 2) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + 6x + 9 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 7x^{3} + 13x^{2} - 3x - 18 = (x + 3)^{2}(x + 2)(x - 1) = 0 \]
\[ x^{4} + 7x^{3} + 13x^{2} - 3x - 18 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & 7 & 13 & -3 & -18 \\ & 0 & 1 & 8 & 21 & 18 \\ \hline & 1 & 8 & 21 & 18 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & 8 & 21 & 18 \\ & 0 & -2 & -12 & -18 \\ \hline & 1 & 6 & 9 & 0 \end{array} \]
\[ x^{2} + 6x + 9 = 0 \Rightarrow x = -3 \]
\[ x^{4} + 7x^{3} + 13x^{2} - 3x - 18 = (x + 3)^{2}(x + 2)(x - 1) \]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 | -7 | -7 |
| Product of the roots | -18 | -18 |
Questions
What are the roots of x⁴ + 7x³ + 13x² - 3x - 18 = 0?
The solutions of x⁴ + 7x³ + 13x² - 3x - 18 = 0 are x = -3, -3, -2, 1 (repeated roots counted once each: -3, -2, 1).
How do you factor x⁴ + 7x³ + 13x² - 3x - 18?
(x + 3)(x + 2)(x − 1).
What is the sum of the roots?
-7 (it equals −7 by Vieta's formulas).