1. ## Another problem child

The problem child I scanned...
(Note: The scanner cut off part of the problem, thats a 4 on the left side there.)

While I work on correcting some problems, this was another one from earlier I was having problems with. Anyone know the right way to solve it?

This should also be the last problem for now I ask for steps on how to do it. (Until I possibly stumble upon another one later...)

I also forgot to post this earlier, its 2 other problems I wanted to check on if I did it right (if I didn't please point out where). Although i'm now noticing I didn't quite finish #2 yet, but am I at least on the right track?

More children with problems...

2. for the 1st problem, start by factoring the numerators and denominators of both fractions ... then do the division by multiplying by the reciprocal of the divisor.

3. = $\frac{4x^2 - 9}{3x^2 + 2x}$ x $\frac{3x^2 - 7x + 6}{2x^2 + 11x + 12}$

= $\frac{(2x)^2 - 3^2}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}$

= $\frac{(2x- 3)(2x + 3)}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}$

= $\frac{2x- 3}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{x + 4}$

could be done further than that.....
$3x^2 - 7x + 6$ cannot be factorized since its discriminant $b^2 - 4ac$ < 0
.....so may be something here is written wrong....

4. Originally Posted by ukorov
= $\frac{4x^2 - 9}{3x^2 + 2x}$ x $\frac{3x^2 - 7x + 6}{2x^2 + 11x + 12}$

= $\frac{(2x)^2 - 3^2}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}$

= $\frac{(2x- 3)(2x + 3)}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}$

= $\frac{2x- 3}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{x + 4}$

could be done further than that.....
$3x^2 - 7x + 6$ cannot be factorized since its discriminant $b^2 - 4ac$ < 0
.....so may be something here is written wrong....
*looks* 3x^2-7x+6 is actually supposed to be 3x^2-7x-6, thats the only part I saw written wrong, could that be it?

Any word about those problems I did?

5. Originally Posted by ukorov
= $\frac{4x^2 - 9}{3x^2 + 2x}$ x $\frac{3x^2 - 7x + 6}{2x^2 + 11x + 12}$

= $\frac{(2x)^2 - 3^2}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}$

= $\frac{(2x- 3)(2x + 3)}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{(2x + 3)(x + 4)}$

= $\frac{2x- 3}{x(3x + 2)}$ x $\frac{3x^2 - 7x + 6}{x + 4}$

could be done further than that.....
$3x^2 - 7x + 6$ cannot be factorized since its discriminant $b^2 - 4ac$ < 0
.....so may be something here is written wrong....
Yes,it's $3x^2-7x-6$ which can be factorized to $(x-3)(3x+2)$