so im home school and its hard to get help on my work,

so im not really looking for an answer to these two questions, but if someone could explain steps how to work it out, itd be very very very much appreciated.

thanks in advance to anybody who helps me.

Have you been taught "long division" in class?
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yes i know regular long division and all that.

but i havent done algebra in almost a year cause ive been doing history and english since june
Its very hard to explain without being able to show you.
Its set out like long division.
First divide 18a^2 by 3a and put it at the top. Copy this down underneath the 18a^2 part Then multiply the part at the top (6a) by 1 (the second part of the polynomial you're dividing by) and put it underneath the 9a part, make sure to inclue the positive/negative sign.
Now draw a line and find the difference. Once you've done that bring down the -5 and do the same until you're finished
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(18a²-9a-5)3a+1)=6a+1+ so that u got now how often can u do 3a in 18a? right 6times
-(18a²-6a) then u multiple 6a with 1 ...its 6a
(3a-5) u still got 3a and -5
-(3a-1)
-4

i dont know the next steps have to think about it....

then u multiple 6a with 1 ...its 6a
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Yeah, pretty much what OddOneOut said. (6a - 5) for the first. And (2r^2 + 6r + 12) for the second.
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^ correct
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this pic should be what the finished result looks like, the method has already been explained...
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Last edited by flambe chicken at Jan 29, 2008,
#1
18a^2 - 9a - 5 / 3a + 1
(6a - 5)(3a + 1) / 3a + 1
(6a - 5) / 1

(6a - 5)
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thanks everyone,

i understand it now, so im pretty confident ill do okay on my chapter test thing today.

so thank you all
Or...you can factorize the first equation.

18a^2 - 9a - 5
= (3a + 1)(6a - 5)

so

(3a + 1)(6a - 5) / (3a +1)

The (3a + 1)s cancel leaving: (6a - 5)
^ Factor the second one mister!
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