1. ## Can someone explain..

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

For all, r ≠ ±2, rČ - 5r + 6 = ?
_______________rČ - 4

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

Thanks!~

2. Originally Posted by anx
I have the answer, but don't understand how it works.

For all, r ≠ ±2, rČ - 5r + 6 = ?
_______________rČ - 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

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

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

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

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

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

I hope this clears it up.

3. 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.

4. 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.

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

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

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

From here we can reduce the fraction by eliminating common factors

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

Do you not know how to factor the expression?

5. 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.

6. 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...

$\displaystyle r^2-5r+6$

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

$\displaystyle ar^2+br+c$

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

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

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

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

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

$\displaystyle (r-3)(r-2)=r^2-2r-3r+6$ when we collect the like terms we end up with $\displaystyle r^2-5r+6$ 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

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

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

Good luck.

7. 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. ><