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Assuming that no denominator equals zero, which expression is equivalent to [tex]\frac{n^2 - 7n + 12}{(n-3)(n+4)}[/tex]?

A. [tex]\frac{n+3}{n-3}[/tex]
B. [tex]\frac{n-4}{n-1}[/tex]
C. [tex]\frac{3(n-1)}{n+4}[/tex]
D. [tex]\frac{n-4}{n+4}[/tex]


Sagot :

Sure, let's solve the given expression step-by-step.

We need to simplify the expression [tex]\(\frac{n^2 - 7n + 12}{(n - 3)(n + 4)}\)[/tex] and determine which of the given options it matches. Here’s the detailed solution:

1. Factor the Numerator:
We start with the numerator [tex]\(n^2 - 7n + 12\)[/tex]. We need to factorize it.
[tex]\[ n^2 - 7n + 12 \][/tex]
To factorize this quadratic expression, we look for two numbers that multiply to [tex]\(12\)[/tex] (the constant term) and add to [tex]\(-7\)[/tex] (the coefficient of [tex]\(n\)[/tex]).

These numbers are [tex]\(-3\)[/tex] and [tex]\(-4\)[/tex], since:
[tex]\[ -3 \times -4 = 12 \][/tex]
[tex]\[ -3 + (-4) = -7 \][/tex]

Therefore, we can write the numerator as:
[tex]\[ n^2 - 7n + 12 = (n - 3)(n - 4) \][/tex]

2. Simplify the Expression:
Now, substitute the factorized form of the numerator back into the original expression:
[tex]\[ \frac{(n - 3)(n - 4)}{(n - 3)(n + 4)} \][/tex]

Notice that both the numerator and the denominator have a common factor, [tex]\((n - 3)\)[/tex]. As long as [tex]\(n \neq 3\)[/tex], we can cancel this common factor:
[tex]\[ \frac{(n - 3)(n - 4)}{(n - 3)(n + 4)} = \frac{n - 4}{n + 4} \][/tex]

3. Result:
After cancelling the common factor, we have:
[tex]\[ \frac{n - 4}{n + 4} \][/tex]

Hence, the simplified form of the given expression [tex]\(\frac{n^2 - 7n + 12}{(n - 3)(n + 4)}\)[/tex] is [tex]\(\frac{n - 4}{n + 4}\)[/tex].

Therefore, the correct option that matches this simplified expression is:
[tex]\[ \boxed{\frac{n - 4}{n + 4}} \][/tex]

Which corresponds to the option:
D. [tex]\(\frac{n - 4}{n + 4}\)[/tex]