Solve x³ + 2x² - 7x + 4 = 0

The solutions of x³ + 2x² - 7x + 4 = 0 are x = -4, 1, 1 (repeated roots counted once each: -4, 1).

x^{3} + 2x^{2} - 7x + 4 = (x + 4)(x - 1)^{2} = 0

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Graph of y = x³ + 2x² - 7x + 4. The curve crosses the x-axis at x = -4, x = 1, 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 4, so the candidates are ±1, ±2, ±4.
$x^{3} + 2x^{2} - 7x + 4 = 0$
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.
$\begin{array}{c|cccc} 1 & 1 & 2 & -7 & 4 \\ & 0 & 1 & 3 & -4 \\ \hline & 1 & 3 & -4 & 0 \end{array}$
3
What is left is the quadratic x² + 3x - 4 = 0. Solve it (factoring or the quadratic formula):
$x^{2} + 3x - 4 = 0 \Rightarrow x = 1,\ -4$
4
So the polynomial factors as
$x^{3} + 2x^{2} - 7x + 4 = (x + 4)(x - 1)^{2}$

LaTeX source of the formulas on this page

\[ x^{3} + 2x^{2} - 7x + 4 = (x + 4)(x - 1)^{2} = 0 \]
\[ x^{3} + 2x^{2} - 7x + 4 = 0 \]
\[ \begin{array}{c|cccc} 1 & 1 & 2 & -7 & 4 \\ & 0 & 1 & 3 & -4 \\ \hline & 1 & 3 & -4 & 0 \end{array} \]
\[ x^{2} + 3x - 4 = 0 \Rightarrow x = 1,\ -4 \]
\[ x^{3} + 2x^{2} - 7x + 4 = (x + 4)(x - 1)^{2} \]

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	-2	-2
Product of the roots	-4	-4

Questions

What are the roots of x³ + 2x² - 7x + 4 = 0?

The solutions of x³ + 2x² - 7x + 4 = 0 are x = -4, 1, 1 (repeated roots counted once each: -4, 1).

How do you factor x³ + 2x² - 7x + 4?

(x + 4)(x − 1).

What is the sum of the roots?

-2 (it equals −2 by Vieta's formulas).

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