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Solve x³ + 10x² + 25x = 0

The solutions of x³ + 10x² + 25x = 0 are x = -5, -5, 0 (repeated roots counted once each: -5, 0).

Graph of y = x³ + 10x² + 25x-6-5-4-3-2-11-20-10102030405060x = -5x = 0y
Graph of y = x³ + 10x² + 25x. The curve crosses the x-axis at x = -5, x = 0, the roots found above.

How to work it out

  1. 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).

    x3+10x2+25x=0x^{3} + 10x^{2} + 25x = 0
  2. 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.

    01102500000110250\begin{array}{c|cccc} 0 & 1 & 10 & 25 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 10 & 25 & 0 \end{array}
  3. 3

    What is left is the quadratic x² + 10x + 25 = 0. Solve it (factoring or the quadratic formula):

    x2+10x+25=0⇒x=−5x^{2} + 10x + 25 = 0 \Rightarrow x = -5
  4. 4

    So the polynomial factors as

    x3+10x2+25x=(x+5)2xx^{3} + 10x^{2} + 25x = (x + 5)^{2}x
LaTeX source of the formulas on this page
\[ x^{3} + 10x^{2} + 25x = (x + 5)^{2}x = 0 \]
\[ x^{3} + 10x^{2} + 25x = 0 \]
\[ \begin{array}{c|cccc} 0 & 1 & 10 & 25 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 10 & 25 & 0 \end{array} \]
\[ x^{2} + 10x + 25 = 0 \Rightarrow x = -5 \]
\[ x^{3} + 10x^{2} + 25x = (x + 5)^{2}x \]

Candidates tested (computed)

Candidate xp(x)Root?
00root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots-10-10
Product of the roots00

Questions

What are the roots of x³ + 10x² + 25x = 0?

The solutions of x³ + 10x² + 25x = 0 are x = -5, -5, 0 (repeated roots counted once each: -5, 0).

How do you factor x³ + 10x² + 25x?

(x + 5)x.

What is the sum of the roots?

-10 (it equals −10 by Vieta's formulas).

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    Solve x³ + 10x² + 25x = 0: x = -5, 0 | WordGPT