Solve x⁴ - 2x³ - 24x² + 18x + 135 = 0
The solutions of x⁴ - 2x³ - 24x² + 18x + 135 = 0 are x = -3, -3, 3, 5 (repeated roots counted once each: -3, 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 135, so the candidates are ±1, ±3, ±5, ±9, ±15, ±27, ±45, ±135.
- 2
Test the candidates until one gives 0. x = 3 is a root. Divide by (x − 3) 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 - 2x - 15 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 2x^{3} - 24x^{2} + 18x + 135 = (x + 3)^{2}(x - 3)(x - 5) = 0 \]
\[ x^{4} - 2x^{3} - 24x^{2} + 18x + 135 = 0 \]
\[ \begin{array}{c|ccccc} 3 & 1 & -2 & -24 & 18 & 135 \\ & 0 & 3 & 3 & -63 & -135 \\ \hline & 1 & 1 & -21 & -45 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 1 & -21 & -45 \\ & 0 & -3 & 6 & 45 \\ \hline & 1 & -2 & -15 & 0 \end{array} \]
\[ x^{2} - 2x - 15 = 0 \Rightarrow x = 5,\ -3 \]
\[ x^{4} - 2x^{3} - 24x^{2} + 18x + 135 = (x + 3)^{2}(x - 3)(x - 5) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 128 | no |
| -1 | 96 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 2 | 2 |
| Product of the roots | 135 | 135 |
Questions
What are the roots of x⁴ - 2x³ - 24x² + 18x + 135 = 0?
The solutions of x⁴ - 2x³ - 24x² + 18x + 135 = 0 are x = -3, -3, 3, 5 (repeated roots counted once each: -3, 3, 5).
How do you factor x⁴ - 2x³ - 24x² + 18x + 135?
(x + 3)(x − 3)(x − 5).
What is the sum of the roots?
2 (it equals −(-2) by Vieta's formulas).