Question

what is the simplest form of \\(\frac{x^2 + 5x - 36}{x^2 - 16}\\)?\
\\(\bigcirc\\) \\(\frac{x + 9}{x + 4}\\)\
\\(\bigcirc\\) \\(\frac{9}{4}\\)\
\\(\bigcirc\\) \\(\frac{x + 9}{x - 4}\\)

Explanation:

Step1: Factor numerator and denominator

Factor \(x^{2}+5x - 36\): We need two numbers that multiply to \(-36\) and add to \(5\). Those numbers are \(9\) and \(-4\), so \(x^{2}+5x - 36=(x + 9)(x-4)\).
Factor \(x^{2}-16\): This is a difference of squares, so \(x^{2}-16=(x + 4)(x - 4)\)? Wait, no, \(x^{2}-16=(x + 4)(x - 4)\)? Wait, \(x^{2}-16=(x + 4)(x - 4)\) is wrong. Wait, \(x^{2}-16=(x + 4)(x - 4)\)? No, \(x^{2}-16=(x + 4)(x - 4)\) is correct? Wait, no, \(x^{2}-16=(x + 4)(x - 4)\) is correct. Wait, but the numerator is \((x + 9)(x - 4)\), denominator is \((x + 4)(x - 4)\)? Wait, no, the original denominator is \(x^{2}-16\), which is \((x + 4)(x - 4)\). Wait, but the numerator is \(x^{2}+5x - 36=(x + 9)(x - 4)\). Then we can cancel the common factor \((x - 4)\) (assuming \(x
eq4\)). So \(\frac{(x + 9)(x - 4)}{(x + 4)(x - 4)}=\frac{x + 9}{x + 4}\)? Wait, no, wait the denominator in the original problem is \(x^{2}-16\), which is \((x + 4)(x - 4)\), and numerator is \((x + 9)(x - 4)\). So cancel \((x - 4)\), we get \(\frac{x + 9}{x + 4}\)? Wait, but the third option is \(\frac{x + 9}{x - 4}\). Wait, maybe I made a mistake in factoring the denominator. Wait, the original problem's denominator is \(x^{2}-16\)? Wait, the user wrote " \(x^{2}-18\)"? Wait, no, looking at the image, the denominator is \(x^{2}-16\)? Wait, the user's text: "What is the simplest form of \(\frac{x^{2}+5x - 36}{x^{2}-18}\)? " No, wait the image shows denominator as \(x^{2}-16\)? Wait, the user's text: " \(x^{2}-18\)"? Wait, maybe it's a typo. But looking at the options, the third option is \(\frac{x + 9}{x - 4}\). Wait, let's re - factor. Wait, if the denominator is \(x^{2}-16=(x + 4)(x - 4)\), numerator \(x^{2}+5x - 36=(x + 9)(x - 4)\). Then cancel \((x - 4)\), we get \(\frac{x + 9}{x + 4}\) (first option). But if the denominator is \(x^{2}-16\), but maybe the original problem's denominator is \(x^{2}-16\). Wait, the first option is \(\frac{x + 9}{x + 4}\), which matches the cancellation.

Wait, let's re - do:

Numerator: \(x^{2}+5x - 36\). Find two numbers: \(a\times b=-36\), \(a + b = 5\). \(9\times(-4)=-36\), \(9+(-4)=5\). So \(x^{2}+5x - 36=(x + 9)(x - 4)\).

Denominator: \(x^{2}-16=(x + 4)(x - 4)\) (difference of squares: \(a^{2}-b^{2}=(a + b)(a - b)\), here \(a = x\), \(b = 4\)).

Now, the fraction is \(\frac{(x + 9)(x - 4)}{(x + 4)(x - 4)}\). Cancel the common factor \((x - 4)\) (for \(x
eq4\)): \(\frac{x + 9}{x + 4}\).

So the first option is \(\frac{x + 9}{x + 4}\), which is the correct simplification.

Step2: Cancel common factors

After factoring, we have \(\frac{(x + 9)(x - 4)}{(x + 4)(x - 4)}\). The common factor \((x - 4)\) can be canceled (as long as \(x
eq4\)), leaving \(\frac{x + 9}{x + 4}\).

Answer:

A. \(\frac{x + 9}{x + 4}\)