Solve x⁴ + 11x³ + 43x² + 69x + 36 = 0
The solutions of x⁴ + 11x³ + 43x² + 69x + 36 = 0 are x = -4, -3, -3, -1 (repeated roots counted once each: -4, -3, -1).
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 = -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 + 7x + 12 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} + 11x^{3} + 43x^{2} + 69x + 36 = (x + 4)(x + 3)^{2}(x + 1) = 0 \]
\[ x^{4} + 11x^{3} + 43x^{2} + 69x + 36 = 0 \]
\[ \begin{array}{c|ccccc} -1 & 1 & 11 & 43 & 69 & 36 \\ & 0 & -1 & -10 & -33 & -36 \\ \hline & 1 & 10 & 33 & 36 & 0 \end{array} \]
\[ \begin{array}{c|cccc} -3 & 1 & 10 & 33 & 36 \\ & 0 & -3 & -21 & -36 \\ \hline & 1 & 7 & 12 & 0 \end{array} \]
\[ x^{2} + 7x + 12 = 0 \Rightarrow x = -3,\ -4 \]
\[ x^{4} + 11x^{3} + 43x^{2} + 69x + 36 = (x + 4)(x + 3)^{2}(x + 1) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 160 | no |
| -1 | 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 | 36 | 36 |
Questions
What are the roots of x⁴ + 11x³ + 43x² + 69x + 36 = 0?
The solutions of x⁴ + 11x³ + 43x² + 69x + 36 = 0 are x = -4, -3, -3, -1 (repeated roots counted once each: -4, -3, -1).
How do you factor x⁴ + 11x³ + 43x² + 69x + 36?
(x + 4)(x + 3)(x + 1).
What is the sum of the roots?
-11 (it equals −11 by Vieta's formulas).