Solve x³ - 11x² + 40x - 48 = 0
The solutions of x³ - 11x² + 40x - 48 = 0 are x = 3, 4, 4 (repeated roots counted once each: 3, 4).
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 -48, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±48.
- 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 - 8x + 16 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - 11x^{2} + 40x - 48 = (x - 3)(x - 4)^{2} = 0 \]
\[ x^{3} - 11x^{2} + 40x - 48 = 0 \]
\[ \begin{array}{c|cccc} 3 & 1 & -11 & 40 & -48 \\ & 0 & 3 & -24 & 48 \\ \hline & 1 & -8 & 16 & 0 \end{array} \]
\[ x^{2} - 8x + 16 = 0 \Rightarrow x = 4 \]
\[ x^{3} - 11x^{2} + 40x - 48 = (x - 3)(x - 4)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -18 | no |
| -1 | -100 | no |
| 2 | -4 | no |
| -2 | -180 | no |
| 3 | 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 | 48 | 48 |
Questions
What are the roots of x³ - 11x² + 40x - 48 = 0?
The solutions of x³ - 11x² + 40x - 48 = 0 are x = 3, 4, 4 (repeated roots counted once each: 3, 4).
How do you factor x³ - 11x² + 40x - 48?
(x − 3)(x − 4).
What is the sum of the roots?
11 (it equals −(-11) by Vieta's formulas).