Solve x⁴ - 11x³ + 42x² - 68x + 40 = 0
The solutions of x⁴ - 11x³ + 42x² - 68x + 40 = 0 are x = 2, 2, 2, 5 (repeated roots counted once each: 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 - 7x + 10 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 11x^{3} + 42x^{2} - 68x + 40 = (x - 2)^{3}(x - 5) = 0 \]
\[ x^{4} - 11x^{3} + 42x^{2} - 68x + 40 = 0 \]
\[ \begin{array}{c|ccccc} 2 & 1 & -11 & 42 & -68 & 40 \\ & 0 & 2 & -18 & 48 & -40 \\ \hline & 1 & -9 & 24 & -20 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 2 & 1 & -9 & 24 & -20 \\ & 0 & 2 & -14 & 20 \\ \hline & 1 & -7 & 10 & 0 \end{array} \]
\[ x^{2} - 7x + 10 = 0 \Rightarrow x = 5,\ 2 \]
\[ x^{4} - 11x^{3} + 42x^{2} - 68x + 40 = (x - 2)^{3}(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 4 | no |
| -1 | 162 | no |
| 2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 11 | 11 |
| Product of the roots | 40 | 40 |
Questions
What are the roots of x⁴ - 11x³ + 42x² - 68x + 40 = 0?
The solutions of x⁴ - 11x³ + 42x² - 68x + 40 = 0 are x = 2, 2, 2, 5 (repeated roots counted once each: 2, 5).
How do you factor x⁴ - 11x³ + 42x² - 68x + 40?
(x − 2)(x − 5).
What is the sum of the roots?
11 (it equals −(-11) by Vieta's formulas).