Solve x⁴ - 4x³ - 13x² + 28x + 60 = 0
The solutions of x⁴ - 4x³ - 13x² + 28x + 60 = 0 are x = -2, -2, 3, 5 (repeated roots counted once each: -2, 3, 5).
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 - 8x + 15 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 4x^{3} - 13x^{2} + 28x + 60 = (x + 2)^{2}(x - 3)(x - 5) = 0 \]
\[ x^{4} - 4x^{3} - 13x^{2} + 28x + 60 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & -4 & -13 & 28 & 60 \\ & 0 & -2 & 12 & 2 & -60 \\ \hline & 1 & -6 & -1 & 30 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & -6 & -1 & 30 \\ & 0 & -2 & 16 & -30 \\ \hline & 1 & -8 & 15 & 0 \end{array} \]
\[ x^{2} - 8x + 15 = 0 \Rightarrow x = 5,\ 3 \]
\[ x^{4} - 4x^{3} - 13x^{2} + 28x + 60 = (x + 2)^{2}(x - 3)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 72 | no |
| -1 | 24 | no |
| 2 | 48 | no |
| -2 | 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 | 60 | 60 |
Questions
What are the roots of x⁴ - 4x³ - 13x² + 28x + 60 = 0?
The solutions of x⁴ - 4x³ - 13x² + 28x + 60 = 0 are x = -2, -2, 3, 5 (repeated roots counted once each: -2, 3, 5).
How do you factor x⁴ - 4x³ - 13x² + 28x + 60?
(x + 2)(x − 3)(x − 5).
What is the sum of the roots?
4 (it equals −(-4) by Vieta's formulas).