this stumps me:

allegansveritatem

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I came across this problem at the end of one section of the algebra review in my precalculus. I will post the problem and how I tried to work it out--it has been asked of us to simplify the expression:
Problem:11369

Here is how I tried to simplify this:
11370

Here is what I got from my calculator:
11371

I really have no idea how the calculator got this and I really have no idea how to simplify this expression. I tried to divide each component of the numerator by the denominator. I mean, what else can you do when neither the numerator or denominator can be fully factored.
 

topsquark

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Hint: Factor the numerator of the original expression...

-Dan
 

Otis

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Can you factor -(r^2 - r - 12) ?

😎
 

pka

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\(\displaystyle \dfrac{12+r-r^2}{r^3+3r^2}\\\dfrac{(4-r)(3+r)}{r^2(r+3)}\\\dfrac{4-r}{r^2}\)
 

allegansveritatem

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\(\displaystyle \dfrac{12+r-r^2}{r^3+3r^2}\\\dfrac{(4-r)(3+r)}{r^2(r+3)}\\\dfrac{4-r}{r^2}\)
Oh boy! I misread the thing...somehow I missed the fact that the numerator is a quadratic expression. What a dope! Thanks for pointing this out. Sometimes it happens...the answer is right there hanging out for all the world to see and...I don't see it. At least I did factor the denominator.
 

allegansveritatem

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Can you factor -(r^2 - r - 12) ?

😎
I see it now but somehow the fact that this is a quadratic expression was completely lost on me. No way to excuse this...I have seen plenty like this with constant out front, but somehow I missed this one. Thanks for replying
 

Otis

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💡 Anytime you're given a rational function (i.e., algebraic fraction), you should first consider factoring everything. That effort is not always needed, but you never know when some exercise might be cooked up for simplification(s).

Cheers 😎
 

topsquark

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I should mention, though it is not always required for full marks, that \(\displaystyle \dfrac{4 - r}{r^2}\) is not quite the correct answer. Since we canceled the r + 3 we need to specify that \(\displaystyle r \neq -3\) for our expression since that is not allowed in the original expression . So the correct answer is actually
\(\displaystyle \dfrac{4 - r}{r^2} \text{ such that } r \neq -3\)

Always look for something like this.

-Dan
 

allegansveritatem

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I went at this again today and here is what I got. It is not quite what the other posters have...but I think that is attributable to some kind of synonomy?11381

Or have I made some kind of mistake? I have 1/R instead of -R/R2 because -R divided by R2 is -1. No?
 

ksdhart2

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I went at this again today and here is what I got. It is not quite what the other posters have...but I think that is attributable to some kind of synonomy?

Or have I made some kind of mistake? I have 1/R instead of -R/R2 because -R divided by R2 is -1. No?
You made a sign error here. Note the negative sign in:

\(\displaystyle \frac{{\color{red} \textbf{-} } R^2 + R + 12}{R^3 + 3R^2}\)

But in the next step, the negative sign has disappeared:

\(\displaystyle \frac{(R + 4)(R + 3)}{R^2(R + 3)}\)

If you re-multiply this out, you'll get \(\displaystyle {\color{red} \textbf{+} }R^2 \cdots\) which should set some alarms blaring.
 

pka

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I went at this again today and here is what I got. It is not quite what the other posters have...but I think that is attributable to some kind of synonomy?

Or have I made some kind of mistake? I have 1/R instead of -R/R2 because -R divided by R2 is -1. No?
\(\displaystyle \dfrac{4-r}{r^2}=\dfrac{4}{r^2}-\dfrac{1}{r} \: (r\ne 0)\)
 

Jomo

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In my opinion, the final answer should be a single fraction. That is although (4/r2) - (1/r) does equal (4-r)/r2, I feel that the final answer should be (4-r)/r2 where r\(\displaystyle \neq\)0
 

pka

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In my opinion, the final answer should be a single fraction. That is although (4/r2) - (1/r) does equal (4-r)/r2, I feel that the final answer should be (4-r)/r2 where r\(\displaystyle \neq\)0
Why do you think that? If this is an online class or an online test then the correct answer is the one expected by the machine and not the one you may prefer.
 

allegansveritatem

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You made a sign error here. Note the negative sign in:

\(\displaystyle \frac{{\color{red} \textbf{-} } R^2 + R + 12}{R^3 + 3R^2}\)

But in the next step, the negative sign has disappeared:

\(\displaystyle \frac{(R + 4)(R + 3)}{R^2(R + 3)}\)

If you re-multiply this out, you'll get \(\displaystyle {\color{red} \textbf{+} }R^2 \cdots\) which should set some alarms blaring.
I don't see what you mean here. I do have a negative sign--although I admit it is not very easy to make out--with (-r+4).
 

Jomo

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Why do you think that? If this is an online class or an online test then the correct answer is the one expected by the machine and not the one you may prefer.
Well first of all we were not given choices. It looks neater when a solution is in factored form as a single fraction. I am sure that you would not expand a denominator if it was in factored form. I guess though that it is a matter of style (if the solutions are not given.
 

ksdhart2

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I don't see what you mean here. I do have a negative sign--although I admit it is not very easy to make out--with (-r+4).
Oh, yes. I see it now that I know it's there. I genuinely thought it was part of the R.
 

Otis

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… -R divided by R2 is -1
That is a typo, yes?

The answer shown in your work is correct (post #10). So is topquark's answer (post #9). These two answer forms are equivalent. As a precalc student, you need to recognize the following.

\(\displaystyle \frac{A - B}{C} = \frac{A}{C} - \frac{B}{C}\)

PS: Don't forget to specify that R ≠ -3 (Your instructor will take note and be pleased!)

Cheers 😎
 

Denis

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I came across this problem at the end of one section of the algebra review in my precalculus. I will post the problem and how I tried to work it out--it has been asked of us to simplify the expression:
Problem:View attachment 11369
Easier if you multiply by -1 right away:
(r^2 - r - 12) / -(r^3 + 3r^2)

Now factor numerator...
 
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