# Rtional Expression...re-submit

• Feb 15th 2008, 04:50 PM
Isdene
Rtional Expression...re-submit
A student was given the following poblem:

4x^2 - 9
___________ * 6x - 9

4x^ + 12 +9

He states his answer to the problem as 3(2x - 3). Is the answer correct? If not, explain why. What would you suggest he do to solve the problem.

The above is the problem as presented to me. I do not have a clue as to how to answer it! Thanks
• Feb 15th 2008, 05:18 PM
mr fantastic
Quote:

Originally Posted by Isdene
A student was given the following poblem:

4x^2 - 9
___________ * 6x - 9

4x^ + 12 +9

Mr F says: I assume that the denominator is 4x^2 + 12x + 9. Although, speculating on the intent of the question, I'll bet it's meant to be 4x^2 - 12x + 9 ......

He states his answer to the problem as 3(2x - 3). Is the answer correct? If not, explain why. What would you suggest he do to solve the problem.

The above is the problem as presented to me. I do not have a clue as to how to answer it! Thanks

If you factorise you get:

$\displaystyle \frac{(2x - 3)(2x + 3)3(2x - 3)}{(2x + 3)^2}$.

You can cancel the (2x + 3) as a common factor provided $\displaystyle 2x + 3 \neq 0 \Rightarrow x \neq -\frac{3}{2}$. Otherwise you're dividing by zero and that's against the rules (on the grounds that everything would be totally insane if you did).

So the answer is $\displaystyle \frac{3(2x - 3)^2}{2x + 3}$, $\displaystyle x \neq -\frac{3}{2}$.

So the student is wrong on two counts:

- they don't have the $\displaystyle x \neq -\frac{3}{2}$ bit.

- they have 3(2x - 3) instead of $\displaystyle \frac{3(2x - 3)^2}{2x + 3}$.

So to solve the problem I'd suggest the student get urgent remedial mathematics help.