anonymous
  • anonymous
I need help finding the restriction on the quotient of a rational expression!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ 6x+12 }{ x+2 }\]
terenzreignz
  • terenzreignz
Shame I don't know what "restriction on the quotient of a rational expression" means -_- However, you CAN simplify that divisor into just a simple number...
anonymous
  • anonymous
Ive already multiplied by the reciprocal and then taken out the GCF of the numerator and the denominator and then factored the denominator leaving me with \[\frac{ 5(x^3-7x-6) }{ 6(x^2+9)(x+2) }\]

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anonymous
  • anonymous
My options are \[x \neq-2\] \[x \neq9\] \[x \neq3\] \[x \neq-3\] and i have to pick the one that is NOT a restriction
terenzreignz
  • terenzreignz
Oh? It's easier than I thought. ^^ There's actually no need for any simplification. Just say, try the first choice: x = -2 Substitute it for the x's in the division and see if it results in anything you might call "illegal"
terenzreignz
  • terenzreignz
And just so you know "illegal" is when you get a zero in the denominator, or a negative number inside a square root sign (irrelevant, since you don't have square roots here)
anonymous
  • anonymous
Thats what i thought, but the only one that i can see IS a restriction is -2
terenzreignz
  • terenzreignz
Good, so the first choice isn't the answer. What about the second? Does replacing the x's with 9's result in any...problems?
anonymous
  • anonymous
nope
terenzreignz
  • terenzreignz
So doesn't that mean the second one is the answer? haha Try the third option. Does substituting 3 for x cause problems?
anonymous
  • anonymous
I thought it would be 9 too but then i tried 3 and that doesn't cause any 0's either
terenzreignz
  • terenzreignz
Are you sure? ;) Look closely at this part: \[\large \frac{ 5x^2-10x-15 }{ \boxed{\color{red}{x^2-9}} } \div \frac{ 6x+12 }{ x+2 }\]
anonymous
  • anonymous
Im supposed to substitute in the original equation???? i thought i was supposed to simplify first....
terenzreignz
  • terenzreignz
You could simplify, but in these cases, original expression is king. And I'll tell you why: If x was, say, -2, then the original equation would have been illegal in the first place, as, therefore, would any simplification of it. So the restrictions have to be binding on the original equation. To be more specific, we could say, generally, I could factor out the 6 here: \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ \color{blue}{6(x+2)} }{ x+2 }\] and then cancel out: \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ 6\cancel{(x+2)} }{\cancel{ x+2 }}\] But this would have been illegal if x was -2. So yeah, use the original expression when considering restrictions ^^
anonymous
  • anonymous
THANK YOU SO MUCH OH MY GOD YOU HAVE NO IDEA HOW MUCH TROUBLE THIS SAVED ME
terenzreignz
  • terenzreignz
No problem ^^ I trust you have no doubts as to your answer now? :D
anonymous
  • anonymous
Nope (; i got it! Thanks so much!!!
terenzreignz
  • terenzreignz
One more thing... and this doesn't really affect THIS particular problem, but I just thought you'd know... since we ARE dividing rational expressions, any value of x that causes the NUMERATOR of the DIVISOR to be equal to zero must also be illegal. Specifically, this part: \[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{\boxed{\color{green}{ 6x+12}} }{ x+2 }\] That's because that also results in a division by zero when you consider the whole thing. Get it? Got it? Good ^^ Good job, by the way :D

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