Question

I need help finding the restriction on the quotient of a rational expression!

Answer 1

\[\frac{ 5x^2-10x-15 }{ x^2-9 } \div \frac{ 6x+12 }{ x+2 }\]

Answer 2

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...

Answer 3

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 \[\frac{ 5(x^3-7x-6) }{ 6(x^2+9)(x+2) }\]

Answer 4

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

Answer 5

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"

Answer 6

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)

Answer 7

Thats what i thought, but the only one that i can see IS a restriction is -2

Answer 8

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?

Answer 9

nope

Answer 10

So doesn't that mean the second one is the answer? haha

Try the third option.
Does substituting 3 for x cause problems?

Answer 11

I thought it would be 9 too but then i tried 3 and that doesn't cause any 0's either

Answer 12

Are you sure? ;)

Look closely at this part:

\[\large \frac{ 5x^2-10x-15 }{ \boxed{\color{red}{x^2-9}} } \div \frac{ 6x+12 }{ x+2 }\]

Answer 13

Im supposed to substitute in the original equation???? i thought i was supposed to simplify first....

Answer 14

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 ^^

Answer 15

THANK YOU SO MUCH OH MY GOD YOU HAVE NO IDEA HOW MUCH TROUBLE THIS SAVED ME

Answer 16

No problem ^^

I trust you have no doubts as to your answer now? :D

Answer 17

Nope (; i got it! Thanks so much!!!

Answer 18

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