Solve x⁴ - 13x² + 12x = 0
The solutions of x⁴ - 13x² + 12x = 0 are x = -4, 0, 1, 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 0 (it is 0, so x = 0 is a root and x factors out).
- 2
Test the candidates until one gives 0. x = 0 is a root. Divide by (x − 0) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
x = 1 is a root. Divide by (x − 1) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 4
What is left is the quadratic x² + x - 12 = 0. Solve it (factoring or the quadratic formula):
- 5
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{4} - 13x^{2} + 12x = (x + 4)x(x - 1)(x - 3) = 0 \]
\[ x^{4} - 13x^{2} + 12x = 0 \]
\[ \begin{array}{c|ccccc} 0 & 1 & 0 & -13 & 12 & 0 \\ & 0 & 0 & 0 & 0 & 0 \\ \hline & 1 & 0 & -13 & 12 & 0 \end{array} \]
\[ \begin{array}{c|cccc} 1 & 1 & 0 & -13 & 12 \\ & 0 & 1 & 1 & -12 \\ \hline & 1 & 1 & -12 & 0 \end{array} \]
\[ x^{2} + x - 12 = 0 \Rightarrow x = 3,\ -4 \]
\[ x^{4} - 13x^{2} + 12x = (x + 4)x(x - 1)(x - 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 0 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 0 | 0 |
| Product of the roots | 0 | 0 |
Questions
What are the roots of x⁴ - 13x² + 12x = 0?
The solutions of x⁴ - 13x² + 12x = 0 are x = -4, 0, 1, 3.
How do you factor x⁴ - 13x² + 12x?
(x + 4)x(x − 1)(x − 3).
What is the sum of the roots?
0 (it equals −0 by Vieta's formulas).