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!~
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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.
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.
$\displaystyle \frac{r^2-5r+6}{r^2-4}$
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?
Please be clear with your questions.
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.
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. ><