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Solve x³ + 5x² + 4x = 0

The solutions of x³ + 5x² + 4x = 0 are x = -4, -1, 0.

Graph of y = x³ + 5x² + 4x-5-4-3-2-11-30-20-101020x = -4x = -1x = 0y
Graph of y = x³ + 5x² + 4x. The curve crosses the x-axis at x = -4, x = -1, 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+5x2+4x=0x^{3} + 5x^{2} + 4x = 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.

    0154000001540\begin{array}{c|cccc} 0 & 1 & 5 & 4 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 5 & 4 & 0 \end{array}
  3. 3

    What is left is the quadratic x² + 5x + 4 = 0. Solve it (factoring or the quadratic formula):

    x2+5x+4=0⇒x=−1, −4x^{2} + 5x + 4 = 0 \Rightarrow x = -1,\ -4
  4. 4

    So the polynomial factors as

    x3+5x2+4x=(x+4)(x+1)xx^{3} + 5x^{2} + 4x = (x + 4)(x + 1)x
LaTeX source of the formulas on this page
\[ x^{3} + 5x^{2} + 4x = (x + 4)(x + 1)x = 0 \]
\[ x^{3} + 5x^{2} + 4x = 0 \]
\[ \begin{array}{c|cccc} 0 & 1 & 5 & 4 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 5 & 4 & 0 \end{array} \]
\[ x^{2} + 5x + 4 = 0 \Rightarrow x = -1,\ -4 \]
\[ x^{3} + 5x^{2} + 4x = (x + 4)(x + 1)x \]

Candidates tested (computed)

Candidate xp(x)Root?
00root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots-5-5
Product of the roots00

Questions

What are the roots of x³ + 5x² + 4x = 0?

The solutions of x³ + 5x² + 4x = 0 are x = -4, -1, 0.

How do you factor x³ + 5x² + 4x?

(x + 4)(x + 1)x.

What is the sum of the roots?

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

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    Solve x³ + 5x² + 4x = 0: x = -4, -1, 0 | WordGPT