anonymous
  • anonymous
Show all work in dividing 9 x squared minus 1 all over 8 x minus 4 divided by the fraction 3 x squared plus 5 x minus 2 all over 2 x squared plus x minus 1 (list restrictions). [Part 2 (2 points)] Use complete sentences to explain your process.
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
so you are given: \[\left(\frac{ 9x^2 -1 }{ 8x - 4 } \right) \div \left( \frac{ 3x^{2}+5x -2 }{ 2x^{2} +x -1 } \right)\]
anonymous
  • anonymous
Dividing by a fraction is the same thing as multiplying by the reciprocal so than you have \[\left( \frac{ 9x^2 -1 }{ 8x -4 } \right) * \left( \frac{ 2x^2 + x - 1 }{ 3x^2 +5 } \right)\] simply multiply straight across to get: \[\left( \frac{ 18x^4 +9x^3 -9x^2 -2x^2 -x +1 }{ 24x^3 +40x -12x^2 -20 } \right)\]
anonymous
  • anonymous
How about just finding the roots of: The denominator of the first divisor The numerator of the dividend The denominator of the dividend

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anonymous
  • anonymous
Next you collect the similar terms \[\left( \frac{ 18x^4 +9x^3 -11x^2 -x + 1 }{ 24x^3 -12 x^2 +40x -20 } \right)\]
anonymous
  • anonymous
@wio ah yeah can just factor the expressions to make division simpiler
anonymous
  • anonymous
The denominators are clear restrictions. The numerator is a restriction because you can't divided by 0.
anonymous
  • anonymous
I'm afraid, though I may be wrong, that doing the actual algebra may get rid of discontinuities that exist, in the same way that \(x/x = 1\) get's rid of the \(x\neq 0\) restrictions.
anonymous
  • anonymous
The whole expression factors to \[\left( \frac{ (x+1)(2x-1)(3x-1)(3x+1) }{ 4(2x-2)(3x^2+5) } \right)\]
anonymous
  • anonymous
now cancel the same terms on top and bottom to simplify to: \[\left( \frac{ (x+1)(3x-1)(3x+1) }{ 4(3x^2 +5) } \right)\]

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