Find the term containing x^5 in the expansion of (2x+3) (x-2)^6

- anonymous

Find the term containing x^5 in the expansion of (2x+3) (x-2)^6

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- anonymous

@zepdrix ^-^

- zepdrix

Hmm I haven't seen this type of problem before :d
I'll have to think about it a sec.

- anonymous

okie no biggie

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## More answers

- 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

alrighty. do you wanna write it all out or should i help (but i think i'd be really bad help tbh)

- 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

ooh i could tell you the answer

- anonymous

and you could work towards getting it

- Astrophysics

shouldn't it be the other way xD

- 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

ahahahah yea probably

- anonymous

but the back of my book has the answers i just don't know how to get them ;-;

- zepdrix

Ok I think we're on the right track then :)
Let's see how this goes.

- zepdrix

Distributing,\[\large\rm =2x \sum(stuff) +3\sum(stuff)\]
Ok with this step?

- anonymous

kinda lol

- anonymous

i mean what's the "stuff" supposed to be?

- zepdrix

The stuff from the orange sum posted before.

- anonymous

ohh ok right right

- 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

\[\large\rm =\sum\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k+1}(-2)^{k+1} +\sum 3\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^{k}\]

- anonymous

im confused :/

- zepdrix

By which part? :o

- anonymous

ok im looking up when you did \[2xSigma(stuff)\]

- 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

also i don't understand where you get the 3

- anonymous

basically everywhere down from there, i don't get

- 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

ohhok

- anonymous

ok ok now i get it

- 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

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

so if you wrote it out, it'd be \[2x * (x^{6-r} * (-2)^r)\]

- anonymous

and then added to what 3 distributed to that stuff would be

- zepdrix

Oh I switched to k again didn't I? woops >.<

- anonymous

haha its fine

- 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

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

ok

- anonymous

so would you set 6-r to 5?

- anonymous

so it'd be \[6-r\]

- anonymous

=5

- 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

ok gotcha

- anonymous

can you give me a minute?

- zepdrix

ya

- zepdrix

imma grab some foods real quick :d

- 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

k=2, r=1.
Ya that sounds right :)

- anonymous

ok but did you see the answer

- anonymous

its kinda....different from what we got

- zepdrix

Is it?
Hmm, looks the same to me.

- 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

wait ok how did they get that from r=2, r=1

- 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

And then we just have to simplify a bit further, ya?

- anonymous

yess

- anonymous

ok then um you would add the 4 to the 2?

- anonymous

but i don't get that either i mean how are they getting 8

- zepdrix

add? 0_o

- anonymous

omg no it's multiplying

- zepdrix

:)

- anonymous

ok you're literally a genius you should get paid for this

- 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

lolol XD aren't you sweet.
i do tutoring in real life sometimes :))
too busy with school'n right now though

- 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

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

oh umm

- anonymous

what? i dont get the last part

- 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

\[\left(\begin{matrix}6 \\ 2\end{matrix}\right)=15,\qquad\qquad maybe\]

- zepdrix

Triangle is used to get those values ^

- anonymous

oh yea

- 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

yea i know how to do that haha

- zepdrix

ok good :)
happy studying!

- zepdrix

and Christmas!

- anonymous

haha thanks you too!

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