anonymous
  • anonymous
Find the term containing x^5 in the expansion of (2x+3) (x-2)^6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@zepdrix ^-^
zepdrix
  • zepdrix
Hmm I haven't seen this type of problem before :d I'll have to think about it a sec.
anonymous
  • anonymous
okie no biggie

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zepdrix
  • zepdrix
I guess we could use Binomial Theorem to turn the second part into a sum, and then distribute that sum to the first set of brackets. maybe.
anonymous
  • anonymous
alrighty. do you wanna write it all out or should i help (but i think i'd be really bad help tbh)
zepdrix
  • zepdrix
\[\large\rm (2x+3)\color{orangered}{(x-2)^6}\quad=(2x+3)\color{orangered}{\sum_{k=0}^{6}\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k}\] Haha I dunno, I don't wanna steal all of the fun XD but it's a little harder to explain if I can't write the stuff out.
anonymous
  • anonymous
ooh i could tell you the answer
anonymous
  • anonymous
and you could work towards getting it
Astrophysics
  • Astrophysics
shouldn't it be the other way xD
anonymous
  • anonymous
\[8\left(\begin{matrix}6 \\ 2\end{matrix}\right) x^5 - 6\left(\begin{matrix}6 \\ 1\end{matrix}\right)x^5 = 84x^5\]
anonymous
  • anonymous
ahahahah yea probably
anonymous
  • anonymous
but the back of my book has the answers i just don't know how to get them ;-;
zepdrix
  • zepdrix
Ok I think we're on the right track then :) Let's see how this goes.
zepdrix
  • zepdrix
Distributing,\[\large\rm =2x \sum(stuff) +3\sum(stuff)\] Ok with this step?
anonymous
  • anonymous
kinda lol
anonymous
  • anonymous
i mean what's the "stuff" supposed to be?
zepdrix
  • zepdrix
The stuff from the orange sum posted before.
anonymous
  • anonymous
ohh ok right right
zepdrix
  • zepdrix
\[\large\rm =2x \sum\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k +3\sum(stuff)\]Let's bring the 2 and x into the first sum, and combine with their respective values,\[\large\rm =\sum\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k+1}(-2)^{k+1} +3\sum(stuff)\] Ok lemme replace the other "stuff" as well at this point.
zepdrix
  • zepdrix
\[\large\rm =\sum\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k+1}(-2)^{k+1} +\sum 3\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^{k}\]
anonymous
  • anonymous
im confused :/
zepdrix
  • zepdrix
By which part? :o
anonymous
  • anonymous
ok im looking up when you did \[2xSigma(stuff)\]
anonymous
  • anonymous
or whatever it was. i just don't get that part. where's the 2x coming from? and what's the "stuff" specifically
anonymous
  • anonymous
also i don't understand where you get the 3
anonymous
  • anonymous
basically everywhere down from there, i don't get
zepdrix
  • zepdrix
We started with Binomial Theorem:\[\large\rm (2x+3)\color{orangered}{(x-2)^6}\quad=(2x+3)\color{orangered}{\sum_{k=0}^{6}\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k}\]We're writing it like this for convenience,\[\large\rm =(2x+3)\color{orangered}{\sum_{k=0}^{6}stuff}\]
anonymous
  • anonymous
ohhok
anonymous
  • anonymous
ok ok now i get it
zepdrix
  • zepdrix
Distribute the sum to each term inside the brackets,\[\large\rm =2x\color{orangered}{\sum_{k=0}^{6}stuff}+3\color{orangered}{\sum_{k=0}^{6}stuff}\]
zepdrix
  • zepdrix
To bring the 2 and x inside the sum, you need to apply exponent rule to each, in order to combine the bases I mean :)
anonymous
  • anonymous
so if you wrote it out, it'd be \[2x * (x^{6-r} * (-2)^r)\]
anonymous
  • anonymous
and then added to what 3 distributed to that stuff would be
zepdrix
  • zepdrix
Oh I switched to k again didn't I? woops >.<
anonymous
  • anonymous
haha its fine
zepdrix
  • zepdrix
For your general term? Ya that sounds right, but with the "choose" bracket thingy also, ya?\[\large\rm =2x\left[\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k\right]+3\left[\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k\right]\]
zepdrix
  • zepdrix
Actually, these k's correspond to separate summations,\[\large\rm =2x\left[\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k\right]+3\left[\left(\begin{matrix}6 \\ r\end{matrix}\right)x^{6-r}(-2)^r\right]\]So I'mma be a little sloppy and use r for the other one. We want to find the value k in the first term that gives us our x^5, and our value of r in the second term that gives us our x^5.
anonymous
  • anonymous
ok
anonymous
  • anonymous
so would you set 6-r to 5?
anonymous
  • anonymous
so it'd be \[6-r\]
anonymous
  • anonymous
=5
zepdrix
  • zepdrix
\[\large\rm =\left[2\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{7-k}(-2)^k\right]+\left[3\left(\begin{matrix}6 \\ r\end{matrix}\right)x^{6-r}(-2)^r\right]\]The first term is a little trickier, because we want to bring the 2 and x in. (I added 1 to the 6-k exponent, that's how he joined in). For the other term? Yes that looks right :) 6-r = 5
anonymous
  • anonymous
ok gotcha
anonymous
  • anonymous
can you give me a minute?
zepdrix
  • zepdrix
ya
zepdrix
  • zepdrix
imma grab some foods real quick :d
anonymous
  • anonymous
okok so then for the first one it'd be r=2 and then the second one would be 1, right? and ok haha
zepdrix
  • zepdrix
k=2, r=1. Ya that sounds right :)
anonymous
  • anonymous
ok but did you see the answer
anonymous
  • anonymous
its kinda....different from what we got
zepdrix
  • zepdrix
Is it? Hmm, looks the same to me.
anonymous
  • anonymous
it's \[8 \left(\begin{matrix}6 \\ 2\end{matrix}\right) x^5 - 6 \left(\begin{matrix}6 \\ 1\end{matrix}\right)x^5 = 84x^5\]
anonymous
  • anonymous
wait ok how did they get that from r=2, r=1
zepdrix
  • zepdrix
\[\large\rm =\left[2\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{7-k}(-2)^k\right]+\left[3\left(\begin{matrix}6 \\ r\end{matrix}\right)x^{6-r}(-2)^r\right]\]We determined that k=2, r=1.\[\large\rm =\left[2\left(\begin{matrix}6 \\ 2\end{matrix}\right)x^{7-2}(-2)^2\right]+\left[3\left(\begin{matrix}6 \\ 1\end{matrix}\right)x^{6-1}(-2)^1\right]\]
zepdrix
  • zepdrix
And then we just have to simplify a bit further, ya?
anonymous
  • anonymous
yess
anonymous
  • anonymous
ok then um you would add the 4 to the 2?
anonymous
  • anonymous
but i don't get that either i mean how are they getting 8
zepdrix
  • zepdrix
add? 0_o
anonymous
  • anonymous
omg no it's multiplying
zepdrix
  • zepdrix
:)
anonymous
  • anonymous
ok you're literally a genius you should get paid for this
anonymous
  • anonymous
my tutor explained it in such a complicated way and did pascal's triangle and everything but this was sooo nice and easy
zepdrix
  • zepdrix
lolol XD aren't you sweet. i do tutoring in real life sometimes :)) too busy with school'n right now though
anonymous
  • anonymous
haha np~ good, i hope you get paid lots lol. anyway, i gotta go study some more but i might pop up later tonight. thanks for all your help!
zepdrix
  • zepdrix
Well you would need to use Pascal's Triangle to do the REST of the problem. Like do you understand how to simply the 6choose2 and 6choose1?
anonymous
  • anonymous
oh umm
anonymous
  • anonymous
what? i dont get the last part
zepdrix
  • zepdrix
\[\left(\begin{matrix}6 \\ 1\end{matrix}\right)=6\]And the other one, I'm not sure, I'd have to go down the triangle to see what the next number in the row is, I think it's 15 or something.
zepdrix
  • zepdrix
\[\left(\begin{matrix}6 \\ 2\end{matrix}\right)=15,\qquad\qquad maybe\]
zepdrix
  • zepdrix
Triangle is used to get those values ^
anonymous
  • anonymous
oh yea
zepdrix
  • zepdrix
Or you can use this if you're comfortable with it:\[\large\rm \left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}\]
anonymous
  • anonymous
yea i know how to do that haha
zepdrix
  • zepdrix
ok good :) happy studying!
zepdrix
  • zepdrix
and Christmas!
anonymous
  • anonymous
haha thanks you too!

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