Question
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Answer
The factored form of the equation is \(3(x+2)(x+4)=0\).
Solution
1. Start with the given equation:
\[
3x^2 + 18x = -24
\]
2. Add \(24\) to both sides to bring all terms to one side:
\[
3x^2 + 18x + 24 = 0
\]
3. Factor out the greatest common factor \(3\):
\[
3(x^2 + 6x + 8) = 0
\]
4. Factor the quadratic \(x^2 + 6x + 8\). We need two numbers that multiply to \(8\) and add up to \(6\). These numbers are \(2\) and \(4\):
\[
x^2 + 6x + 8 = (x+2)(x+4)
\]
5. Substitute the factored form back into the equation:
\[
3(x+2)(x+4) = 0
\]
Thus, the factored form of the equation is:
\[
3(x+2)(x+4)=0
\]
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Simplify this solution Bonus Knowledge
Let's start by rearranging the equation to set it to zero: \( 3x^{2} + 18x + 24 = 0 \) Next, we can factor out the common factor of 3 from the entire equation: \( 3(x^{2} + 6x + 8) = 0 \) Now, we need to focus on factoring the quadratic \( x^{2} + 6x + 8 \). We are looking for two numbers that multiply to 8 and add to 6. These numbers are 2 and 4. Thus, we can factor the quadratic as: \( 3(x + 2)(x + 4) = 0 \) Setting each factor equal to zero gives us the solutions: \( x + 2 = 0 \) or \( x + 4 = 0 \) So, the final solutions are: \( x = -2 \) and \( x = -4 \).
