anonymous
  • anonymous
f(x) = x2 - 16 and g(x) = x+4. Find F over G of and its domain
Mathematics
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anonymous
  • anonymous
f(x) = x2 - 16 and g(x) = x+4. Find F over G of and its domain
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
\[\frac{ f(x) }{ g(x) }\] is this what you mean?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
\[\frac{ (x^2 -16) }{ (x+4) }\] you can factor the numerator in a way so as to get two expressions one of which will be (x+4) which will cancel out with the denominator

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anonymous
  • anonymous
so in other words (x+4)(?) = (x^2 - 16)
anonymous
  • anonymous
4x and 16x^2 right
anonymous
  • anonymous
im sorry i don't quite understand your expression can you write it out?
anonymous
  • anonymous
im not getting it
anonymous
  • anonymous
okay do you know how to factor \[(x^2-16)\]
welshfella
  • welshfella
x^2 - 16 is the difference of 2 squares x^2 is a perfect square and so is 4
welshfella
  • welshfella
have you factored something like that before?
anonymous
  • anonymous
no
anonymous
  • anonymous
are you familiar with the difference of squares method (its the way to factor expresions like these )
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anonymous
  • anonymous
yeah but its a real number right
anonymous
  • anonymous
basically it means that if your constant (in this case the 16) is a perfect square you can factor the expression by multiplying two expressions which's constant is the perfect square of your original constant
anonymous
  • anonymous
okay im over complicating this XD
anonymous
  • anonymous
okay do you know what \[\sqrt{16} = ?\]
anonymous
  • anonymous
2*8
anonymous
  • anonymous
umm no , what number squared equals 16
anonymous
  • anonymous
256
anonymous
  • anonymous
no thats 16 squared
welshfella
  • welshfella
no what number when multiplied by itself equals 16 like 3 * 3 = 9 So 3 is the square root of 9.
anonymous
  • anonymous
4
welshfella
  • welshfella
right
anonymous
  • anonymous
correct which means that \[\sqrt{16} = 4\] and because 4 is a whole number we call 16 a perfect square
anonymous
  • anonymous
okay thanks

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