Solve x³ - 14x² + 64x - 96 = 0
The solutions of x³ - 14x² + 64x - 96 = 0 are x = 4, 4, 6 (repeated roots counted once each: 4, 6).
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 -96, so the candidates are ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±16, ±24, ±32, ±48, ±96.
- 2
Test the candidates until one gives 0. x = 4 is a root. Divide by (x − 4) 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
\[ x^{3} - 14x^{2} + 64x - 96 = (x - 4)^{2}(x - 6) = 0 \]
\[ x^{3} - 14x^{2} + 64x - 96 = 0 \]
\[ \begin{array}{c|cccc} 4 & 1 & -14 & 64 & -96 \\ & 0 & 4 & -40 & 96 \\ \hline & 1 & -10 & 24 & 0 \end{array} \]
\[ x^{2} - 10x + 24 = 0 \Rightarrow x = 6,\ 4 \]
\[ x^{3} - 14x^{2} + 64x - 96 = (x - 4)^{2}(x - 6) \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | -45 | no |
| -1 | -175 | no |
| 2 | -16 | no |
| -2 | -288 | no |
| 3 | -3 | no |
| -3 | -441 | no |
| 4 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 14 | 14 |
| Product of the roots | 96 | 96 |
Questions
What are the roots of x³ - 14x² + 64x - 96 = 0?
The solutions of x³ - 14x² + 64x - 96 = 0 are x = 4, 4, 6 (repeated roots counted once each: 4, 6).
How do you factor x³ - 14x² + 64x - 96?
(x − 4)(x − 6).
What is the sum of the roots?
14 (it equals −(-14) by Vieta's formulas).