Solve x⁴ + 6x³ - 3x² - 52x - 60 = 0
The solutions of x⁴ + 6x³ - 3x² - 52x - 60 = 0 are x = -5, -2, -2, 3 (repeated roots counted once each: -5, -2, 3).
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 -60, so the candidates are ±1, ±2, ±3, ±4, ±5, ±6, ±10, ±12, ±15, ±20, ±30, ±60.
- 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
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 + 2x - 15 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 6x^{3} - 3x^{2} - 52x - 60 = (x + 5)(x + 2)^{2}(x - 3) = 0 \]
\[ x^{4} + 6x^{3} - 3x^{2} - 52x - 60 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & 6 & -3 & -52 & -60 \\ & 0 & -2 & -8 & 22 & 60 \\ \hline & 1 & 4 & -11 & -30 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & 4 & -11 & -30 \\ & 0 & -2 & -4 & 30 \\ \hline & 1 & 2 & -15 & 0 \end{array} \]
\[ x^{2} + 2x - 15 = 0 \Rightarrow x = 3,\ -5 \]
\[ x^{4} + 6x^{3} - 3x^{2} - 52x - 60 = (x + 5)(x + 2)^{2}(x - 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -108 | no |
| -1 | -16 | no |
| 2 | -112 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -6 | -6 |
| Product of the roots | -60 | -60 |
Questions
What are the roots of x⁴ + 6x³ - 3x² - 52x - 60 = 0?
The solutions of x⁴ + 6x³ - 3x² - 52x - 60 = 0 are x = -5, -2, -2, 3 (repeated roots counted once each: -5, -2, 3).
How do you factor x⁴ + 6x³ - 3x² - 52x - 60?
(x + 5)(x + 2)(x − 3).
What is the sum of the roots?
-6 (it equals −6 by Vieta's formulas).