Question

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

Answer 1

@zepdrix ^-^

Answer 2

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

Answer 3

okie no biggie

Answer 4

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.

Answer 5

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

Answer 6

\[\large\rm (2x+3)\color{orangered}{(x-2)^6}\quad=(2x+3)\color{orangered}{\sum_{k=0}^{6}\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k}\]

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.

Answer 7

ooh i could tell you the answer

Answer 8

and you could work towards getting it

Answer 9

shouldn't it be the other way xD

Answer 10

\[8\left(\begin{matrix}6 \\ 2\end{matrix}\right) x^5 - 6\left(\begin{matrix}6 \\ 1\end{matrix}\right)x^5 = 84x^5\]

Answer 11

ahahahah yea probably

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

kinda lol

Answer 16

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

Answer 17

The stuff from the orange sum posted before.

Answer 18

ohh ok right right

Answer 19

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

Answer 20

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

Answer 21

im confused :/

Answer 22

By which part? :o

Answer 23

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

Answer 24

or whatever it was. i just don't get that part. where's the 2x coming from? and what's the "stuff" specifically

Answer 25

also i don't understand where you get the 3

Answer 26

basically everywhere down from there, i don't get

Answer 27

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,\[\large\rm =(2x+3)\color{orangered}{\sum_{k=0}^{6}stuff}\]

Answer 28

ohhok

Answer 29

ok ok now i get it

Answer 30

Distribute the sum to each term inside the brackets,\[\large\rm =2x\color{orangered}{\sum_{k=0}^{6}stuff}+3\color{orangered}{\sum_{k=0}^{6}stuff}\]

Answer 31

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 :)

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

haha its fine

Answer 36

For your general term?
Ya that sounds right, but with the "choose" bracket thingy also, ya?\[\large\rm =2x\left[\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k\right]+3\left[\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{6-k}(-2)^k\right]\]

Answer 37

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.

Answer 38

ok

Answer 39

so would you set 6-r to 5?

Answer 40

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

Answer 41

=5

Answer 42

\[\large\rm =\left[2\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{7-k}(-2)^k\right]+\left[3\left(\begin{matrix}6 \\ r\end{matrix}\right)x^{6-r}(-2)^r\right]\]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

Answer 43

ok gotcha

Answer 44

can you give me a minute?

Answer 45

ya

Answer 46

imma grab some foods real quick :d

Answer 47

okok so then for the first one it'd be r=2 and then the second one would be 1, right? and ok haha

Answer 48

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

Answer 49

ok but did you see the answer

Answer 50

its kinda....different from what we got

Answer 51

Is it?
Hmm, looks the same to me.

Answer 52

it's \[8 \left(\begin{matrix}6 \\ 2\end{matrix}\right) x^5 - 6 \left(\begin{matrix}6 \\ 1\end{matrix}\right)x^5 = 84x^5\]

Answer 53

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

Answer 54

\[\large\rm =\left[2\left(\begin{matrix}6 \\ k\end{matrix}\right)x^{7-k}(-2)^k\right]+\left[3\left(\begin{matrix}6 \\ r\end{matrix}\right)x^{6-r}(-2)^r\right]\]We determined that k=2, r=1.\[\large\rm =\left[2\left(\begin{matrix}6 \\ 2\end{matrix}\right)x^{7-2}(-2)^2\right]+\left[3\left(\begin{matrix}6 \\ 1\end{matrix}\right)x^{6-1}(-2)^1\right]\]

Answer 55

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

Answer 56

yess

Answer 57

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

Answer 58

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

Answer 59

add? 0_o

Answer 60

omg no it's multiplying

Answer 61

:)

Answer 62

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

Answer 63

my tutor explained it in such a complicated way and did pascal's triangle and everything but this was sooo nice and easy

Answer 64

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

Answer 65

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!

Answer 66

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?

Answer 67

oh umm

Answer 68

what? i dont get the last part

Answer 69

\[\left(\begin{matrix}6 \\ 1\end{matrix}\right)=6\]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.

Answer 70

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

Answer 71

Triangle is used to get those values ^

Answer 72

oh yea

Answer 73

Or you can use this if you're comfortable with it:\[\large\rm \left(\begin{matrix}n \\ r\end{matrix}\right)=\frac{n!}{r!(n-r)!}\]

Answer 74

yea i know how to do that haha

Answer 75

ok good :)
happy studying!

Answer 76

and Christmas!

Answer 77

haha thanks you too!