Solve x³ - 5x² - 8x + 12 = 0

The solutions of x³ - 5x² - 8x + 12 = 0 are x = -2, 1, 6.

x^{3} - 5x^{2} - 8x + 12 = (x + 2)(x - 1)(x - 6) = 0

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Graph of y = x³ - 5x² - 8x + 12. The curve crosses the x-axis at x = -2, x = 1, x = 6, the roots found above.

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 12, so the candidates are ±1, ±2, ±3, ±4, ±6, ±12.
$x^{3} - 5x^{2} - 8x + 12 = 0$
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.
$\begin{array}{c|cccc} 1 & 1 & -5 & -8 & 12 \\ & 0 & 1 & -4 & -12 \\ \hline & 1 & -4 & -12 & 0 \end{array}$
3
What is left is the quadratic x² - 4x - 12 = 0. Solve it (factoring or the quadratic formula):
$x^{2} - 4x - 12 = 0 \Rightarrow x = 6,\ -2$
4
So the polynomial factors as
$x^{3} - 5x^{2} - 8x + 12 = (x + 2)(x - 1)(x - 6)$

LaTeX source of the formulas on this page

\[ x^{3} - 5x^{2} - 8x + 12 = (x + 2)(x - 1)(x - 6) = 0 \]
\[ x^{3} - 5x^{2} - 8x + 12 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & -5 & -8 & 12 \\ & 0 & 1 & -4 & -12 \\ \hline & 1 & -4 & -12 & 0 \end{array} \]
\[ x^{2} - 4x - 12 = 0 \Rightarrow x = 6,\ -2 \]
\[ x^{3} - 5x^{2} - 8x + 12 = (x + 2)(x - 1)(x - 6) \]

Candidates tested (computed)

Candidate x	p(x)	Root?
1	0	root

Vieta's formulas check the answer (computed)

Relation	From the roots	From the coefficients
Sum of the roots	5	5
Product of the roots	-12	-12

Questions

What are the roots of x³ - 5x² - 8x + 12 = 0?

The solutions of x³ - 5x² - 8x + 12 = 0 are x = -2, 1, 6.

How do you factor x³ - 5x² - 8x + 12?

(x + 2)(x − 1)(x − 6).

What is the sum of the roots?

5 (it equals −(-5) by Vieta's formulas).

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