Solve x⁴ - 2x³ - 11x² + 12x + 36 = 0
The solutions of x⁴ - 2x³ - 11x² + 12x + 36 = 0 are x = -2, -2, 3, 3 (repeated roots counted once each: -2, 3).
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 36, so the candidates are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
- 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 - 6x + 9 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 2x^{3} - 11x^{2} + 12x + 36 = (x + 2)^{2}(x - 3)^{2} = 0 \]
\[ x^{4} - 2x^{3} - 11x^{2} + 12x + 36 = 0 \]
\[ \begin{array}{c|ccccc} -2 & 1 & -2 & -11 & 12 & 36 \\ & 0 & -2 & 8 & 6 & -36 \\ \hline & 1 & -4 & -3 & 18 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -2 & 1 & -4 & -3 & 18 \\ & 0 & -2 & 12 & -18 \\ \hline & 1 & -6 & 9 & 0 \end{array} \]
\[ x^{2} - 6x + 9 = 0 \Rightarrow x = 3 \]
\[ x^{4} - 2x^{3} - 11x^{2} + 12x + 36 = (x + 2)^{2}(x - 3)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 36 | no |
| -1 | 16 | no |
| 2 | 16 | no |
| -2 | 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 | 36 | 36 |
Questions
What are the roots of x⁴ - 2x³ - 11x² + 12x + 36 = 0?
The solutions of x⁴ - 2x³ - 11x² + 12x + 36 = 0 are x = -2, -2, 3, 3 (repeated roots counted once each: -2, 3).
How do you factor x⁴ - 2x³ - 11x² + 12x + 36?
(x + 2)(x − 3).
What is the sum of the roots?
2 (it equals −(-2) by Vieta's formulas).