Can someone explain..

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June 4th 2008, 11:52 AM
anx

Can someone explain..

I have the answer, but don't understand how it works.

For all, r ≠ ±2, r² - 5r + 6 = ?
_______________r² - 4

r - 3
r + 2 is the answer, but I have no idea how I am supposed to get it.

Thanks!~
June 4th 2008, 11:57 AM
TheEmptySet

Quote:

Originally Posted by anx

I have the answer, but don't understand how it works.

For all, r ≠ ±2, r² - 5r + 6 = ?
_______________r² - 4

r - 3
r + 2 is the answer, but I have no idea how I am supposed to get it.

Thanks!~

We start by factoring the numerator and the denominator

$\frac{r^2-5r+6}{r^2-4}=\frac{(r-2)(r-3)}{(r-2)(r+2)}$

Now we can reduce out the common factor of (r-2) to get

$\frac{r-3}{r+2}$

Note: that even though the factor r-2 reduced out $r \ne 2$

becuase we would still be dividing by zero in the original equation.

I hope this clears it up.
June 4th 2008, 12:03 PM
anx

Thank you, but that is not exactly the way for me to figure it out.

Say I don't have the answer in the first place, and simply need to answer the original question, how would i find

r - 3
r + 2

I only said i had the answer in the first place because I am doing a practice test which includes the answers, which usually lets me figure out things backwards, but in this case I do not know how, and certainly will not have the answer to do so on the real test.
June 4th 2008, 12:21 PM
TheEmptySet

Quote:

Originally Posted by anx

Thank you, but that is not exactly the way for me to figure it out.

Say I don't have the answer in the first place, and simply need to answer the original question, how would i find

r - 3
r + 2

I only said i had the answer in the first place because I am doing a practice test which includes the answers, which usually lets me figure out things backwards, but in this case I do not know how, and certainly will not have the answer to do so on the real test.

I didn't start with the answer I started with the expression.

$\frac{r^2-5r+6}{r^2-4}$

If we factor this, as in my above post, we get

$\frac{(r-2)(r-3)}{(r-2)(r+2)}$

From here we can reduce the fraction by eliminating common factors

$\frac{(r-3)}{(r+2)}$

Do you not know how to factor the expression?

Please be clear with your questions.
June 4th 2008, 12:32 PM
anx

Quote:

Originally Posted by TheEmptySet

Do you not know how to factor the expression?

I suppose not. This is probably why I don't understand it. All I see is 0/0 every time.
June 4th 2008, 12:42 PM
TheEmptySet

Quote:

Originally Posted by anx

I suppose not. This is probably why I don't understand it. All I see is 0/0 every time.

I think I see your problem:

We are not using the values r=2 or r=-2.

In fact they are not in the domain of the function so we are not allowed to use them.

lets look at the numerator for a second...

$r^2-5r+6$

We need to factor this expression loosely speaking un Distribute or FOIL it

To do this we need to compare it to the general form

$ar^2+br+c$

we can identify a=1 b=-5 and c=6.

Since a=1 when just need to find factors of c (6) that add up to b (-5)

We start by looking at all of the factors of 6 they are

$1\cdot 6 \\\ 2\cdot 3$

We know that since c is postitive and b is negative that both of our factors need to be negative.

$(-2)(-3)=6 \\\ (-2)+(-3)=-5$ So it seems that these work so we can factor $r^2-5r+6=(r-3)(r-2)$

We can check that the factorization is correct by multiplying it out and seeing if it equals what we started with so lets check

$(r-3)(r-2)=r^2-2r-3r+6$ when we collect the like terms we end up with $r^2-5r+6$ so our factorization checks.

Note: that the denominator can be factored this way, but it is in a special form called the difference of squares that can always be factored using this formula

$a^2-b^2=(a+b)(a-b)$

Hint: for yours remember that $4=2^2$

Good luck.
June 4th 2008, 12:54 PM
anx

Well I understand what you did now. Hopefully I can apply it when I need to, but I don't have another sample problem like that to try it. ><