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Solve x³ + 10x² + 24x = 0

The solutions of x³ + 10x² + 24x = 0 are x = -6, -4, 0.

Graph of y = x³ + 10x² + 24x-7-6-5-4-3-2-11-202040x = -6x = -4x = 0y
Graph of y = x³ + 10x² + 24x. The curve crosses the x-axis at x = -6, x = -4, 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+24x=0x^{3} + 10x^{2} + 24x = 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.

    01102400000110240\begin{array}{c|cccc} 0 & 1 & 10 & 24 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 10 & 24 & 0 \end{array}
  3. 3

    What is left is the quadratic x² + 10x + 24 = 0. Solve it (factoring or the quadratic formula):

    x2+10x+24=0⇒x=−4, −6x^{2} + 10x + 24 = 0 \Rightarrow x = -4,\ -6
  4. 4

    So the polynomial factors as

    x3+10x2+24x=(x+6)(x+4)xx^{3} + 10x^{2} + 24x = (x + 6)(x + 4)x
LaTeX source of the formulas on this page
\[ x^{3} + 10x^{2} + 24x = (x + 6)(x + 4)x = 0 \]
\[ x^{3} + 10x^{2} + 24x = 0 \]
\[ \begin{array}{c|cccc} 0 & 1 & 10 & 24 & 0 \\ & 0 & 0 & 0 & 0 \\ \hline & 1 & 10 & 24 & 0 \end{array} \]
\[ x^{2} + 10x + 24 = 0 \Rightarrow x = -4,\ -6 \]
\[ x^{3} + 10x^{2} + 24x = (x + 6)(x + 4)x \]

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 + 24x = 0?

The solutions of x³ + 10x² + 24x = 0 are x = -6, -4, 0.

How do you factor x³ + 10x² + 24x?

(x + 6)(x + 4)x.

What is the sum of the roots?

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

Similar equations

    Solve x³ + 10x² + 24x = 0: x = -6, -4, 0 | WordGPT