anonymous
  • anonymous
Multiplying rational expressions: x+5/4x-16 * 2x^2-32/x^2-25
Mathematics
jamiebookeater
  • jamiebookeater
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ash2326
  • ash2326
we have \[\frac{x+5}{4x+16} * \frac{2x^2-32}{x^2-25}\] now we can take out 4 from 4x +16 and 2 from 2x^2-32 so we get \[\frac{x+5}{4(x+4)} * \frac{2(x^2-16)}{x^2-25}\] now using a^-b^2=(a+b)(a-b) \[\frac{x+5}{4(x+4)} * \frac{2(x+4)(x-4)}{(x-5)(x+5)}\] now we can cancel x+5 and x+4 from numerator and denominator, so we have \[\frac{\cancel{x+5}}{4{\cancel{(x+4)}}} * \frac{2\cancel{(x+4)}(x-4)}{(x-5)\cancel{(x+5)}}\] so we have now \[\frac{2(x-4)}{4(x-5)}\] we get finally \[\frac{x-4}{2(x-5)}\]
anonymous
  • anonymous
why would we take oout 4 and 2?
anonymous
  • anonymous
I will tell you if i said .. i give 2 apple for 4 persons then of course each person have \[\frac{1}{2}\] apple its the simplest form while you said it \[\frac{2}{4}, \frac{1}{2}\] both are true ... The origin of Your Qustion is simplification .. so ash2326 do that in best way .

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