Solve x⁴ + 7x³ + 6x² - 28x - 40 = 0
The solutions of x⁴ + 7x³ + 6x² - 28x - 40 = 0 are x = -5, -2, -2, 2 (repeated roots counted once each: -5, -2, 2).
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 -40, so the candidates are ±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40.
- 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 + 7x + 10 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 7x^{3} + 6x^{2} - 28x - 40 = (x + 5)(x + 2)^{2}(x - 2) = 0 \]
\[ x^{4} + 7x^{3} + 6x^{2} - 28x - 40 = 0 \]
\[ \begin{array}{c|ccccc} 2 & 1 & 7 & 6 & -28 & -40 \\ & 0 & 2 & 18 & 48 & 40 \\ \hline & 1 & 9 & 24 & 20 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & 9 & 24 & 20 \\ & 0 & -2 & -14 & -20 \\ \hline & 1 & 7 & 10 & 0 \end{array} \]
\[ x^{2} + 7x + 10 = 0 \Rightarrow x = -2,\ -5 \]
\[ x^{4} + 7x^{3} + 6x^{2} - 28x - 40 = (x + 5)(x + 2)^{2}(x - 2) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -54 | no |
| -1 | -12 | no |
| 2 | 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 | -40 | -40 |
Questions
What are the roots of x⁴ + 7x³ + 6x² - 28x - 40 = 0?
The solutions of x⁴ + 7x³ + 6x² - 28x - 40 = 0 are x = -5, -2, -2, 2 (repeated roots counted once each: -5, -2, 2).
How do you factor x⁴ + 7x³ + 6x² - 28x - 40?
(x + 5)(x + 2)(x − 2).
What is the sum of the roots?
-7 (it equals −7 by Vieta's formulas).