1. ## Stuck

I am way stuck on trying to perform the indicated operations and simplify this expression: ( 6r^2 - 7r - 3)/( r^2 - 1) divided by (4r^2 - 12r +9) /( r^2 - 1) . (2r^2 - r -3/ 3r^2 - 2r -1) ????

2. Originally Posted by jay1
I am way stuck on trying to perform the indicated operations and simplify this expression: ( 6r^2 - 7r - 3)/( r^2 - 1) divided by (4r^2 - 12r +9) /( r^2 - 1) . (2r^2 - r -3/ 3r^2 - 2r -1) ????
Hi jay1,

$\displaystyle \frac{6r^2-7r-3}{r^2-1} \div \frac{4r^2-12r+9}{r^2-1} \cdot \frac{2r^2-r-3}{3r^2-2r-1}$

First, invert the divisor and multiply and cancel out the $\displaystyle r^2-1$ terms.

Next, factor all you can and simplify. Let's see what you can do!

3. Originally Posted by masters
Hi jay1,

$\displaystyle \frac{6r^2-7r-3}{r^2-1} \div \frac{4r^2-12r+9}{r^2-1} \cdot \frac{2r^2-r-3}{3r^2-2r-1}$

First, invert the divisor and multiply and cancel out the $\displaystyle r^2-1$ terms.

Next, factor all you can and simplify. Let's see what you can do!
Is it r-1/r+1??

5. $\displaystyle \frac{(2r-3)(3r+1)}{(2r-3)^2}\ \frac{(r+1)(2r-3)}{(r-1)(3r+1)} = \frac{(r+1)}{(r-1)}$

as masters suggested you may not of inverted the divide

6. When I divide this expression, I'm coming up with a remainder. Is that right? The expression is 27z + 23z^2 + 6z^3 divided by 2z + 3

7. Originally Posted by jay1
When I divide this expression, I'm coming up with a remainder. Is that right? The expression is 27z + 23z^2 + 6z^3 divided by 2z + 3
$\displaystyle (6z^3+23z^2+27z) \div (2z+3)$

Use polynomial long division.

$\displaystyle 3z^2+7z+\frac{6z}{2z+3}$