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Solve x³ + 3x² - 4 = 0

The solutions of x³ + 3x² - 4 = 0 are x = -2, -2, 1 (repeated roots counted once each: -2, 1).

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Graph of y = x³ + 3x² - 4-3-2-112-10-5510152025x = -2x = 1y
Graph of y = x³ + 3x² - 4. The curve crosses the x-axis at x = -2, x = 1, 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 -4, so the candidates are ±1, ±2, ±4.

    x3+3x2−4=0x^{3} + 3x^{2} - 4 = 0
  2. 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.

    1130−401441440\begin{array}{c|cccc} 1 & 1 & 3 & 0 & -4 \\ & 0 & 1 & 4 & 4 \\ \hline & 1 & 4 & 4 & 0 \end{array}
  3. 3

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

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

    So the polynomial factors as

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

Candidates tested (computed)

Candidate xp(x)Root?
10root

Vieta's formulas check the answer (computed)

RelationFrom the rootsFrom the coefficients
Sum of the roots-3-3
Product of the roots44

Questions

What are the roots of x³ + 3x² - 4 = 0?

The solutions of x³ + 3x² - 4 = 0 are x = -2, -2, 1 (repeated roots counted once each: -2, 1).

How do you factor x³ + 3x² - 4?

(x + 2)(x − 1).

What is the sum of the roots?

-3 (it equals −3 by Vieta's formulas).

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    Solve x³ + 3x² - 4 = 0: x = -2, 1 | WordGPT