Solve x³ + 7x² - 6x - 72 = 0
The solutions of x³ + 7x² - 6x - 72 = 0 are x = -6, -4, 3.
How to work it out
- 1
A cubic has up to 3 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 -72, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±36, ±72.
- 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
What is left is the quadratic x² + 10x + 24 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} + 7x^{2} - 6x - 72 = (x + 6)(x + 4)(x - 3) = 0 \]
\[ x^{3} + 7x^{2} - 6x - 72 = 0 \]
\[ \begin{array}{c|cccc} 3 & 1 & 7 & -6 & -72 \\ & 0 & 3 & 30 & 72 \\ \hline & 1 & 10 & 24 & 0 \end{array} \]
\[ x^{2} + 10x + 24 = 0 \Rightarrow x = -4,\ -6 \]
\[ x^{3} + 7x^{2} - 6x - 72 = (x + 6)(x + 4)(x - 3) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -70 | no |
| -1 | -60 | no |
| 2 | -48 | no |
| -2 | -40 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | -7 | -7 |
| Product of the roots | 72 | 72 |
Questions
What are the roots of x³ + 7x² - 6x - 72 = 0?
The solutions of x³ + 7x² - 6x - 72 = 0 are x = -6, -4, 3.
How do you factor x³ + 7x² - 6x - 72?
(x + 6)(x + 4)(x − 3).
What is the sum of the roots?
-7 (it equals −7 by Vieta's formulas).