- Factor the numerator \(3x^2-3\) to \(3(x-1)(x+1)\) and the denominator \(x^2+3x\) to \(x(x+3)\).
- Multiply the numerator by the reciprocal of the denominator: \(\frac{3(x-1)(x+1)}{x(x+3)} \cdot \frac{x(x+3)}{x+1}\).
- Cancel out the common factors \(x\), \((x+3)\), and \((x+1)\) from the numerator and denominator.
- Simplify the remaining terms to reach the final answer of \(\boxed{3(x - 1)}\).
Explanation
- Analysis of the Expression
We are asked to simplify the algebraic expression:
\[ \frac{\frac{3x^2 - 3}{x^2 + 3x}}{\frac{x + 1}{x(x + 3)}} \]
To simplify this quotient of two rational expressions, we first factor the numerator and denominator wherever possible and then divide by multiplying by the reciprocal. The domain of the expression is restricted to values of \(x\) where denominators are not zero: \(x
eq 0\), \(x
eq -3\), and \(x
eq -1\).
- Factoring the Numerator
First, we focus on the numerator of the main fraction:
\[ \frac{3x^2 - 3}{x^2 + 3x} \]
We can factor the numerator \(3x^2 - 3\) as \(3(x^2 - 1)\), which is a difference of squares: \(3(x - 1)(x + 1)\).
The denominator \(x^2 + 3x\) factors as \(x(x + 3)\).
Thus, the numerator of the main fraction is:
\[ \frac{3(x - 1)(x + 1)}{x(x + 3)} \]
- Multiplying by the Reciprocal
Now we rewrite the entire quotient as a product by multiplying the numerator of the main fraction by the reciprocal of the denominator:
\[ \left( \frac{3(x - 1)(x + 1)}{x(x + 3)} \right) \times \left( \frac{x(x + 3)}{x + 1} \right) \]
By combining these fractions, we get:
\[ \frac{3(x - 1)(x + 1) \cdot x(x + 3)}{x(x + 3) \cdot (x + 1)} \]
- Simplifying the Expression
Now, we identify and cancel the common factors in the numerator and the denominator:
- The common factor \(x\) exists in both.
- The common factor \((x + 3)\) exists in both.
- The common factor \((x + 1)\) exists in both.
After canceling these common factors, we are left with:
\[ 3(x - 1) \]
This expression is equivalent to the given quotient. Comparing this result with the provided choices, it matches the second option.
Examples
Algebraic simplification is widely used in fields like physics and engineering when dealing with complex formulas. For instance, when calculating the velocity of a particle or the resistance in an electrical circuit, you often end up with complex fractional expressions. Simplifying these expressions is crucial for performing further analysis or solving for a specific variable, such as time \(t\) or current \(I\), efficiently. Similar to this problem, we factor and cancel out terms to make our calculations manageable and to gain insights into how the system behaves.