anonymous
  • anonymous
What is the simplified form of x^2 - 16 / x + 4
Mathematics
katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
Hint: x^2 - 16 x^2 - 4^2 (x-4)(x+4)
anonymous
  • anonymous
x - 4, with the restriction x ≠ 4 x + 4, with the restriction x ≠ - 4 x - 4, with the restriction x ≠ - 4 x + 4, with the restriction x ≠ 4
jim_thompson5910
  • jim_thompson5910
I'm using the difference of squares rule

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anonymous
  • anonymous
@jim_thompson5910 is it A?
jim_thompson5910
  • jim_thompson5910
it is x-4, but the restriction is not x ≠ 4 since 4 is a perfectly valid input in the original expression
jim_thompson5910
  • jim_thompson5910
x+4 = 0 x = -4 is the value that makes the denominator zero, so this is the restricted value
anonymous
  • anonymous
So it would be C
jim_thompson5910
  • jim_thompson5910
yes, it's C
anonymous
  • anonymous
@jim_thompson5910 can u plz explain me more i'm confused
jim_thompson5910
  • jim_thompson5910
which part are you confused about?
jim_thompson5910
  • jim_thompson5910
if you want, you can ask the question in a separate post
anonymous
  • anonymous
all
jim_thompson5910
  • jim_thompson5910
\[\Large \frac{x^2 - 16}{x+4}\] \[\Large \frac{x^2 - 4^2}{x+4}\] \[\Large \frac{(x-4)(x+4)}{x+4}\] \[\Large \frac{(x-4)\cancel{(x+4)}}{\cancel{x+4}}\] \[\Large x-4\] So \[\Large \frac{x^2 - 16}{x+4}\] simplifies to \[\Large x-4\] Keep in mind that x cannot equal -4 in the original expression. So for the two expressions to be completely equivalent, x cannot equal -4 in the final expression.

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