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Simplify completely x2+4x-5 . 20x-12 -------- ------- 5x2-8x+3 x2-6x-55
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x^2 + 4x - 5 = (x + 5)(x - 1) 20x - 12 = 4(5x - 3) 5x^2 - 8x + 3 = (5x - 3)(x - 1) x^2 - 6x - 55 = (x - 11)(x + 5) when we write in fractions (x+5)(x-1)*4(5x-3) / (5x - 3)(x -1)(x -11)(x +5) all gets cancelled and we get 4/(x-11)
thank you!! :)
Factor all quadratics \[\frac{x^2+4x-5 }{ 5x^2-8x+3 }\frac{20x-12}{ x^2-6x-55}\] \[\frac{(x-1)(x+5) }{ (x-1)(5x-3) }\frac{4(5x-3)}{ (x-11)(x+5)}\] \[\frac{\cancel{(x-1)}\cancel{(x+5)} }{ \cancel{ (x-1)}\cancel{(5x-3)} }\frac{4\cancel{(5x-3)}}{ (x-11)\cancel{(x+5)}}\] \[\frac{4}{x-11}\]
thanks sam! :)
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