Asked by sayletyler

Multiply x^2-9 over x+2 (3x+6 over x+3)
Also side note can you keep the expalnation nice and short please

Answers

Answered by sayletyler
Ok thats not even close to the size I want way too big
Answered by sayletyler
Ok brotha I need you to size it down more like 4 setences at most
Answered by GPT-4o mini
To multiply the expressions \((x^2 - 9) \div (x + 2)\) and \(\frac{3x + 6}{x + 3}\), we can proceed as follows:

1. **Factor where possible**:
- \(x^2 - 9\) is a difference of squares:
\[
x^2 - 9 = (x - 3)(x + 3)
\]
- \(3x + 6\) can be factored as:
\[
3x + 6 = 3(x + 2)
\]

2. **Rewrite the expression**:
\[
\frac{(x - 3)(x + 3)}{(x + 2)} \cdot \frac{3(x + 2)}{(x + 3)}
\]

3. **Cancel common factors**:
- The \((x + 2)\) in the numerator and denominator cancels.
- The \((x + 3)\) in the numerator and denominator cancels.

4. **Resulting expression**:
After canceling, we are left with:
\[
3(x - 3)
\]

5. **Final answer**:
\[
3(x - 3) = 3x - 9
\]

So, the product is \(3x - 9\).
Answered by GPT-4o mini
Got it! Here’s a much shorter response:

Multiply:

\[
\frac{x^2 - 9}{x + 2} \cdot \frac{3x + 6}{x + 3}
\]

1. Factor:
- \(x^2 - 9 = (x - 3)(x + 3)\)
- \(3x + 6 = 3(x + 2)\)

2. Rewrite:
\[
\frac{(x - 3)(x + 3)}{(x + 2)} \cdot \frac{3(x + 2)}{(x + 3)}
\]

3. Cancel common factors:
- \((x + 2)\) cancels
- \((x + 3)\) cancels

Result:
\[
3(x - 3) = 3x - 9
\]

Final answer: **\(3x - 9\)**.
Answered by GPT-4o mini
To multiply \(\frac{x^2 - 9}{x + 2} \cdot \frac{3x + 6}{x + 3}\):

1. Factor: \((x^2 - 9) = (x - 3)(x + 3)\) and \(3x + 6 = 3(x + 2)\).
2. Rewrite as \(\frac{(x - 3)(x + 3)}{(x + 2)} \cdot \frac{3(x + 2)}{(x + 3)}\).
3. Cancel \((x + 2)\) and \((x + 3)\).
4. Result: \(3(x - 3) = 3x - 9\).

Final answer: **\(3x - 9\)**.

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