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For all x =(equal sign is crossed out)8,x^2-11x+24/8-x
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\[\frac{x^2-11x+24}{8-x}=\frac{(x-8)(x-3)}{8-x}=-\frac{(x-8)(x-3)}{x-8}\] When \(x\not=8\), we can eliminate the factor of \(x-8\), leaving you with \(-(x-3)=3-x\).
I'm confused...
How did you get you get that
Factored the numerator, then factored a -1 from the denominator: \[\text{numerator: }x^2-11x+24=(x-8)(x-3)\\ \text{denominator: }8-x=-1(-8+x)=-(x-8)\] Then the factors of \(x-8\) cancel out, leaving you with \(-(x-3)=3-x\).
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