Solve x³ + x² = 0

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

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

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Graph of y = x³ + x². The curve crosses the x-axis at x = -1, x = 0, 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 0 (it is 0, so x = 0 is a root and x factors out).
$x^{3} + x^{2} = 0$
2
Test the candidates until one gives 0. x = 0 is a root. Divide by (x − 0) with synthetic division; the bottom row is the quotient and a 0 remainder.
$\begin{array}{c|cccc} 0 & 1 & 1 & 0 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 1 & 0 & 0 \end{array}$
3
What is left is the quadratic x² + x = 0. Solve it (factoring or the quadratic formula):
$x^{2} + x = 0 \Rightarrow x = 0,\ -1$
4
So the polynomial factors as
$x^{3} + x^{2} = (x + 1)x^{2}$

LaTeX source of the formulas on this page

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

Candidates tested (computed)

Candidate x	p(x)	Root?
0	0	root

Vieta's formulas check the answer (computed)

Relation	From the roots	From the coefficients
Sum of the roots	-1	-1
Product of the roots	0	0

Questions

What are the roots of x³ + x² = 0?

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

How do you factor x³ + x²?

(x + 1)x.

What is the sum of the roots?

-1 (it equals −1 by Vieta's formulas).

Similar equations

x³ - x = 0