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