find the area of the rectangle with a length of 2x+2/x-4 and the width of 3x-12/x^2+6x+5

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We're trying to evaluate a fractional expression here: [latex]\frac{2x + 2}{x -4} * \frac{3x - 12}{x^{2} + 6x + 5}[/latex] The first thing to do, is factor everything we can. The quadratic factors to (x + 1)(x+5) and the expression above that factors to 3(x - 4). 2x + 2 factors to 2(x + 1). [latex]\frac{2(x + 1)}{x -4} * \frac{3(x - 4)}{(x+1)(x+5)}[/latex] It looks a little complicated now, but if we combine the two we can cancel some stuff. [latex]\frac{2(x + 1) * 3(x - 4)}{(x -4)(x+1)(x+5)}[/latex] Still looking complicated, but look, we have (x+1) and (x+4) on the top and the bottom, that means they don't matter and we can get rid of them: [latex]\frac{2 * 3}{x+5} = \frac{6}{x+5}[/latex] Tada! The area is 6/(x+5). We'd need to know x to go any further! Need any more explanation?

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