Question

question
solve algebraically for all values of x:
$-x^3 - 4x^2 + 3x + 12 = 0$
answer attempt 1 out of 2
additional solution no solution
$x = $
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Explanation:

Step1: Factor by grouping

First, we can factor out a negative sign from the first two terms and the last two terms:
$$-x^3 - 4x^2 + 3x + 12 = 0$$
$$-x^2(x + 4) + 3(x + 4) = 0$$

Step2: Factor out the common binomial

Now, we can factor out the common binomial \((x + 4)\):
$$(x + 4)(-x^2 + 3) = 0$$

Step3: Set each factor equal to zero

We set each factor equal to zero and solve for \(x\):

1. For \(x + 4 = 0\), we get \(x = -4\).
2. For \(-x^2 + 3 = 0\), we can rewrite it as \(x^2 = 3\), so \(x = \pm\sqrt{3}\).

Answer:

\(x = -4\), \(x = \sqrt{3}\), \(x = -\sqrt{3}\)