Question

Simplify
9r^3-4rs^2/2s^2-5sr+3r^2

What's the answer?

Answer 1

\[\large \frac{9r^3-4rs^2}{2s^2-5sr+3r^2}\]Let's factor the top part of the fraction first. Factoring out an r gives us,\[\huge \frac{\color{#F35633}{r}(9r^2-4s^2)}{2s^2-5sr+3r^2}\]Now on top we can see that we have the DIFFERENCE OF SQUARES. Remember how to break down the difference of squares? They can be written as conjugates.\[\huge \frac{\color{#F35633}{r}\color{ #3366CF}{(3r-2s)(3r+2s)}}{2s^2-5sr+3r^2}\]Understand what I did to the top? Or need me to explain that a little better? :)

Answer 2

Oh i understand what u did there :3

Answer 3

So on the bottom.. it's a little tricky to see, but it can be factored.
I'm not sure of the best way to explain it, I just kinda fiddled around with the numbers and guessed the factors.\[\large (2s-3r)(s-r)\]

Answer 4

Is that the final answer? r(3r-2s)(3r+2s)/(2s-3r)(s-r)?

Answer 5

Giving us,\[\huge \frac{r(3r-2s)(3r+2s)}{(2s-3r)(s-r)}\]Factoring a negative out of the term with subtraction gives us, \[\huge \frac{-r(2s-3r)(3r+2s)}{(2s-3r)(s-r)}
\]And it looks like we can do a little cancellation from here.

Answer 6

\[\huge \frac{-r\cancel{(2s-3r)}(3r+2s)}{\cancel{(2s-3r)}(s-r)}\]

Answer 7

Oh i see , so that would be the final ?

Answer 8

-r(3r+2s)/s-r is the dinal?

Answer 9

Hmm maybe there was an easier way to do that :(
I don't think we can simplify it any further.
We could throw the negative into the denominator, giving us,\[\huge \frac{r(3r+2s)}{(r-s)}\]Meaning that we probably should have factored a little differently :d Since we factored out a negative and then put it back in.. that was silly.. my bad.

I THINK this is as far as we can go :D Hopefully...

Answer 10

Final**

Answer 11

Oh thank you so much!

Answer 12

I checked it on a website, yup looks like we did it correctly :)

Answer 13

Btw how did -r become positive and the s-r become r-s?

Answer 14

Let's pretend it was \(-r(s-r)\) for a moment.
If we distribute the negative to the terms inside the brackets, it gives us \(r(-s+r)\) which we can write as \(r(r-s)\)

We used that type of idea, except the (s-r) term was in the bottom.

Answer 15

Oh i see, thanks!