Solve x⁴ - 4x³ - 22x² + 100x - 75 = 0
The solutions of x⁴ - 4x³ - 22x² + 100x - 75 = 0 are x = -5, 1, 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 -75, so the candidates are ±1, ±3, ±5, ±15, ±25, ±75.
- 2
Test the candidates until one gives 0. x = 1 is a root. Divide by (x − 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² - 25 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 4x^{3} - 22x^{2} + 100x - 75 = (x + 5)(x - 1)(x - 3)(x - 5) = 0 \]
\[ x^{4} - 4x^{3} - 22x^{2} + 100x - 75 = 0 \]
\[ \begin{array}{c|ccccc} 1 & 1 & -4 & -22 & 100 & -75 \\ & 0 & 1 & -3 & -25 & 75 \\ \hline & 1 & -3 & -25 & 75 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 3 & 1 & -3 & -25 & 75 \\ & 0 & 3 & 0 & -75 \\ \hline & 1 & 0 & -25 & 0 \end{array} \]
\[ x^{2} - 25 = 0 \Rightarrow x = 5,\ -5 \]
\[ x^{4} - 4x^{3} - 22x^{2} + 100x - 75 = (x + 5)(x - 1)(x - 3)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 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 | -75 | -75 |
Questions
What are the roots of x⁴ - 4x³ - 22x² + 100x - 75 = 0?
The solutions of x⁴ - 4x³ - 22x² + 100x - 75 = 0 are x = -5, 1, 3, 5.
How do you factor x⁴ - 4x³ - 22x² + 100x - 75?
(x + 5)(x − 1)(x − 3)(x − 5).
What is the sum of the roots?
4 (it equals −(-4) by Vieta's formulas).