anonymous
  • anonymous
What is the simplified form of x plus 2 over x squared minus 3x minus 10 • x minus 3 over x squared plus x minus 12 ? 1 over the quantity x minus 3 times the quantity x plus 4 1 over the quantity x minus 3 times the quantity x plus 2 1 over the quantity x plus 4 times the quantity x minus 5 1 over the quantity x plus 2 times the quantity x minus 5
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ x + 2 }{ x^2 - 3x - 10 } \times \frac{ x - 3 }{ x^2 + x -12 }\]
anonymous
  • anonymous
@iGreen do i flip the 2nd fraction then horizontally multiply ?
iGreen
  • iGreen
No, that was for division..you can horizontally multiply now.

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anonymous
  • anonymous
oh okay
anonymous
  • anonymous
quick question when we horizontally multipl would the numerator be x^2 + 2 -3 or x+ 2 x- 3?
iGreen
  • iGreen
None.. \(\sf (x+2)(x-3)\) x * x = ? x * -3 = ? 2 * x = ? 2 * -3 = ?
anonymous
  • anonymous
x^2 -3x 2 -6
iGreen
  • iGreen
Actually, first we cancel the common factors.
anonymous
  • anonymous
for the denominator ?
iGreen
  • iGreen
For the 2nd fraction
iGreen
  • iGreen
1st*
iGreen
  • iGreen
Factor \(\sf x^2 - 3x - 10\) from the numerator..tell me what you get.
iGreen
  • iGreen
Can you do that?
anonymous
  • anonymous
is it x/ x^2 - 3x -5
iGreen
  • iGreen
No..
iGreen
  • iGreen
Can you factor \(\sf x^2 - 3x - 10\)?
anonymous
  • anonymous
oh yeah give me sec
iGreen
  • iGreen
Okay
anonymous
  • anonymous
(x + 2) (x - 5)
iGreen
  • iGreen
Yes, so we have: \(\sf \dfrac{x+2}{(x-5)(x+2)} \times \dfrac{x-3}{x^2+x-12}\) The x + 2's cancel out, giving us: \(\sf \dfrac{\cancel{x+2}}{(x-5)(\cancel{x+2})} \times \dfrac{x-3}{x^2+x-12}\) \(\sf \dfrac{1}{(x-5)} \times \dfrac{x-3}{x^2+x-12}\)
iGreen
  • iGreen
Now we multiply horizontally..but first factor \(\sf x^2 + x - 12\).
anonymous
  • anonymous
(x - 3) (x + 4)
anonymous
  • anonymous
but it would be \[\frac{ 1 }{ (x - 5) } \times \frac{ 1 } { ( x + 4)}\]
iGreen
  • iGreen
Yes..which gives us: \(\sf \dfrac{1}{(x-5)(x+4)}\)
iGreen
  • iGreen
anonymous
  • anonymous
okay thanks so much for some reason i lost connection thats why im replying late

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