Solve x⁴ - x³ - 31x² + 25x + 150 = 0
The solutions of x⁴ - x³ - 31x² + 25x + 150 = 0 are x = -5, -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 150, so the candidates are ±1, ±2, ±3, ±5, ±6, ±10, ±15, ±25, ±30, ±50, ±75, ±150.
- 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 = 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
\[ x^{4} - x^{3} - 31x^{2} + 25x + 150 = (x + 5)(x + 2)(x - 3)(x - 5) = 0 \]
\[ x^{4} - x^{3} - 31x^{2} + 25x + 150 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & -1 & -31 & 25 & 150 \\ & 0 & -2 & 6 & 50 & -150 \\ \hline & 1 & -3 & -25 & 75 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 3 & 1 & -3 & -25 & 75 \\ & 0 & 3 & 0 & -75 \\ \hline & 1 & 0 & -25 & 0 \end{array} \]
\[ x^{2} - 25 = 0 \Rightarrow x = 5,\ -5 \]
\[ x^{4} - x^{3} - 31x^{2} + 25x + 150 = (x + 5)(x + 2)(x - 3)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 144 | no |
| -1 | 96 | no |
| 2 | 84 | no |
| -2 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 1 | 1 |
| Product of the roots | 150 | 150 |
Questions
What are the roots of x⁴ - x³ - 31x² + 25x + 150 = 0?
The solutions of x⁴ - x³ - 31x² + 25x + 150 = 0 are x = -5, -2, 3, 5.
How do you factor x⁴ - x³ - 31x² + 25x + 150?
(x + 5)(x + 2)(x − 3)(x − 5).
What is the sum of the roots?
1 (it equals −(-1) by Vieta's formulas).