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\[\frac{ x^2-5x+6 }{ x^-9 }=\frac{ (x-3)(x-2) }{ (x-3)(x+3) }\]

that's an x^2-9 at the bottom btw, my mistake

no i need a knew one not the one i posted that was an example

Oh, does it need to have all different numbers?

yea i think so

it can be easy it just has to have varibles and needs to be able to be simplified

The previous? Or first?

previous the answer

ok im gonna turn it in ill tell u if your right

she said we have to simplify and get a final answer @Luigi0210

so its right so far were just not done

Well I showed it in the deleted post :l

i know i copy and pasted she said we wernt done

(x-2)(x-3)/(x-2)(x+2)=x^2-5x+6/x^2-4

\[\frac{ x-3 }{ x+2 }\]