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the first one is the equation your simplifying the next 3 are the options :)
Eh, I would just factorise the first term, then multiply by the inverse of \[n+4 / n-3\]
So the first term reduces to (n+4)^2/(n-3)^2 after factorising then you just multiply that by (n-3)/(n+4) which cancels the first to (n+4)/(n-3)
Ahhh i see i would have gone with C i wish there were a quicker easier way to find the answers to these things that werent so dang complicated!! :P lol but thanks a bunch just helped me slot!! :D
your welcome :D
so then the answer to this one would be : wait here i will attach them can u do the last two with me? pls? :)
OK so factorise (n^2 + 3n + 2) to (n+2)(n+1), so it becomes \[((n+2)(n+1))\div2\] then multiply that by the reciprocal of (n+1)/(2(n+2)) which is 2(n+2) / (n+1) , both the 2s cancel and the (n+1)s cancel so its just (n+2)^2
Ahhhh alright i see the "pattern" or whatever lol okie dokie last one and its on long division UGH :P i stil cant get the hang of it :P
OK well can you do long division? Kinda hard to show it on here but I can try
:P i hate long division URGHHHH
ok so do you think you could attach the steps?
Lol x - 9 ______________ x+7/x^2 - 2x - 30 x^2 + 7x -------- -9x - 30 -9x - 63 ---------- 33 <- remainder so it's x - 9 + 33/(x+7)
Ok im pretty sure i get it now, well more than i got it before :) lol thanks :)
i always get so wrapped up in the solving underneath the division thing i forget about above it ^^
haha!! thx!! :)
Yeah.. I remember doing long division of polynomials last term, it's annoying cause I never used to use long division in lower school I just did the divisions in my head so I had no idea how to do long division. Good luck and you're welcome..
yea ikr lol thx :)