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What is the simplified form of x plus 3 over x squared minus x minus 12 • x minus 4 over x squared minus 8x plus 16 ?

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\[\frac{ x+3 }{ x^2-x-12 }*\frac{ x-4 }{ x^2-8x+16 }\]
First you need to factorize both denominators. i would show you how to factorize the denominator, I would take \(x^2 - x -12\) as the example and then you have to proceed yourself.

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Other answers:

I have \(x^2 - x - 12\) \( x^2 + 3x - 4x -12\) = \(x(x+3) - 4(x+3)\) \((x-4)(x+3)\) This is how I factorized \(x^2-x-12\) by using "middle term splitting" method. Similarly, factorize \(x^2 - 8x + 16\) by splitting -8x into -4x and -4x : \(x^2 - 4x -4x + 16\) and then do the proceedings.
You will have then : \(\cfrac{x+3}{(x-4)(x+3)} \times \cfrac{(x-4)}{x^2-8x+16}\) \(\cfrac{(x+3)(x-4)}{(x-4)(x+3)(x^2-8x+16)}\) \(\cfrac{\cancel{(x+3)(x-4)}}{\cancel{(x-4)(x+3)} (x^2-8x+16)}\) \(\cfrac{1}{x^2-8x+16}\) put the factorized form of \(x^2-8x+16\) on the place of it and you will get the required answer. I hope it helped. Post here if you get any problem Do not forget to tell what you get.. best of luck!
I got \[\frac{ 1 }{ (x - 4)^2 }\]is that correct?

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