- Factor the numerator \(x^2+8x-48\) into \((x+12)(x-4)\) and the denominator \(x^2-144\) into \((x+12)(x-12)\).
- Rewrite the division as multiplication by \(\frac{1}{4-x} = \frac{1}{-(x-4)}\).
- Cancel the common factors \((x+12)\) and \((x-4)\) from the numerator and denominator.
- Obtain the final simplified result \(\boxed{\frac{-1}{x-12}}\).
Explanation
- Factoring the Expression
We are tasked with simplifying the expression \(\frac{x^2+8x-48}{x^2-144} \div (4-x)\). To do this, we should factor each part of the expression and then perform the division.
First, let's factor the numerator of the first fraction, \(x^2+8x-48\). We look for two numbers that multiply to \(-48\) and add to \(8\). Those numbers are \(12\) and \(-4\). Thus, \(x^2+8x-48 = (x+12)(x-4)\).
Next, we factor the denominator, \(x^2-144\). This is a difference of squares: \(x^2 - 12^2 = (x-12)(x+12)\).
The original expression is:
\[\frac{(x+12)(x-4)}{(x-12)(x+12)} \div (4-x)\]
- Simplifying the Expression
Now, we simplify the division by rewriting it as multiplication. Dividing by \((4-x)\) is the same as multiplying by \(\frac{1}{4-x}\).
Recall that \(4-x = -(x-4)\). Substituting this, the expression becomes:
\[\frac{(x+12)(x-4)}{(x-12)(x+12)} \cdot \frac{1}{-(x-4)}\]
Now, we can cancel common terms. The \((x+12)\) terms in the numerator and denominator cancel out, as do the \((x-4)\) terms (leaving a factor of \(-1\) due to the negative sign in the denominator).
\[\frac{1 \cdot (x-4)}{(x-12) \cdot 1} \cdot \frac{1}{-(x-4)} = \frac{x-4}{(x-12)(-(x-4))}\]
Canceling \((x-4)\), we are left with:
\[\frac{1}{(x-12)(-1)} = \frac{1}{-(x-12)} = \frac{-1}{x-12}\]
- Conclusion and Final Answer
The expression simplifies to \(\frac{-1}{x-12}\). Comparing this to the given options, it matches the first option. Therefore, the simplest form is \(\frac{-1}{x-12}\).
Examples
Simplifying rational expressions is essential in algebra when analyzing functions or solving complex equations. For example, if you are designing a filter for an audio signal processing system, you might encounter transfer functions that are ratios of polynomials. Simplifying these ratios allows you to identify critical frequencies (poles and zeros) more easily, helping you understand how the system amplifies or attenuates different signals. By breaking down complex fractions, you gain a clearer view of the underlying mathematical behavior.