Question

Expand and simplify (sqrt 3 + 2)^5. Give your answer in the form a+bsqrt

Answer 1

\[(\sqrt{3} + 2)^{5}\]

Answer 2

I know pascals triangle is going to make this 1, 5, 10, 5, 1 and how to order all the exponents and such. my work ended up looking like the following

Answer 3

\[\sqrt{3}^{5} + 5(\sqrt{3})^{4}(2)^{2} + 10(\sqrt{3})^3(2)^2 + 10(\sqrt{3})^2(2)^3 + 5(\sqrt{3})(2)^4 + (2)^5\]

Answer 4

@freckles @whpalmer4 @UsukiDoll any of you guys awake and in the mood to (possibly) help out? I understand if not, it's p late ;-;

Answer 5

hi so um the answer in the back of the book is \[362+209\sqrt{3}\] but i got

Answer 6

\[521+1562\] \[521+1562\sqrt{3}\]

Answer 7

(the last one, the site just glitched and kept the first answer)

Answer 8

ahh, is it too difficult/long ;-;?

Answer 9

\[(a+b)^5 = a^5b^0 + a^4b^1+a^3b^2 + a^2b^3 + a^1b^4+a^0b^5\]with the appropriate coefficients, right?
Why don't we simplify this part first?
\[a^5 = \sqrt{3}^5 = 9\sqrt{5}\]\[a^4 = 9\]\[a^3=3\sqrt{3}\]\[a^2 = 3\]\[a^1 = \sqrt{3}\]
\[b^5 = 32\]\[b^4=16\]\[b^3=8\]\[b^2=4\]\[b^1=2\]

so keeping our term places together so we can stick in the coefficients, that's

\[9\sqrt{3}*1 + 9*2+3\sqrt{3}*4 + 3*8 + \sqrt{3}*16+32\]and sticking in the coefficients from the triangle
\[=1*9\sqrt{3}*1 + 5*9*2 + 10*3\sqrt{3}*4+10*3*8 + 5*\sqrt{3}*16+1*32\]
\[=9\sqrt{3} + 90 + 120\sqrt{3} +240+80\sqrt{3}+32 = (9+120+80)\sqrt{3} + 90+240+32\]\[=362+209\sqrt{3} \]

Answer 10

ah man thank you so much!! makes sense now that I look at it. do you think you could help me with umm one more?

Answer 11

I am a huge fan of doing this kind of tedious work with simple things like \(a\) instead of \(\sqrt{3}\)!

Sure, I could probably stumble through another one...

Answer 12

ok haha thank you. so basically, find the coefficient of \[x^-6\] in the expansion of \[(2x-\frac{ 3 }{ x^2})\]

Answer 13

this one just messes me up cos of the negative exponent

Answer 14

Isn't there supposed to be an exponent on that?

Answer 15

oh yea haha it's 12

Answer 16

could you explain this the way you did with the last example?

Answer 17

darn. that was looking pretty easy :-)

Answer 18

hehe sorry

Answer 19

Okay, well, we have to figure out the coefficients from the triangle. Can you do that?

\[a^{12}b^0 ,a^{11}b^1, a^{10}b^2, a^9b^3, a^8b^4, a^7b^5 ,a^6b^6, a^5b^7, a^4b^8 ,a^3b^9, a^2b^{10} ,a^1b^{11} ,a^0b^{12}\]

Answer 20

Now we have \[a= 2x\]\[b=-3x^{-2}\]Right? And we want to find the coefficient of \(x^{-6}\)

(by the way, if your exponent takes up multiple characters, like "-6", wrap the entire exponent in { } so that the right thing happens)

Answer 21

do you want me to do pascals triangle for 12?

Answer 22

Yes, please. Division of labor :-)

Answer 23

lol kk one sec

Answer 24

1 12 66 220 495 792 924 792 495 220 66 12 1

Answer 25

there's actually a formula that you can use to find the value of any of the coefficients, but I don't ever remember it!

Answer 26

i don't know it either :/

Answer 27

Yes, those the ones.

Now can you tell me which term is going to have the \(x^{-6}\) in it?

Answer 28

umm...792?

Answer 29

is it the second 792?

Answer 30

well, in terms of a^n b^m

by the way, the formula for it is\[\left(\begin{matrix}n \\ k\end{matrix}\right)= \frac{n!}{k!(n-k)!}\]

Answer 31

oh yaaa the binomial theorem

Answer 32

but how can that be used to find the coefficient here? can we do so by finding the general term?

Answer 33

The first term is \(a^{12}b^0\) which is going to \(x^{12}*1\)
The second term is \(a^{11}b^1\) which is going to have \(x^{11}*x^{-2} = x^9\)
The third term is \(a^{10}b^2\) which is going to have \(x^{10}*x^{-2*2} = x^6\)
so the exponent is decreasing by 3 with each term...that means x^3, x^0, x^-3,x^-6 7th term is the one we want

that will be \(a^6b^6 = (2x)^6(-3x^-2)^6 = \)
times our coefficient which is 924

Answer 34

ok

Answer 35

oops, I muffed some formatting.
7th term is \(924a^6b^6 = 924(2x)^6(-3x^{-2})^6 = \)
can you do that part?

Answer 36

yes but it's 1 am where i am and i need 2 to go to bed soon
so i guess we can stop here cos i get the gist of it ^-^

Answer 37

but thank u so much for ur help!!! i really appreciate it<3 have a nice night and thank u for all ur amazing help :')

Answer 38

\[924(2x)^6(-3x^{-2})^6 = 924*2^6x^6*(-3)^6(x^{-2})^6\]\[ = 924*64*729*x^6*x^{-12} \]\[= 43110144x^{-6}\]

Answer 39

(thank u)

Answer 40

the whole thing is \[\frac{531441}{x^{24}}-\frac{4251528}{x^{21}}+\frac{15588936}{x^{18}}-\frac{34642080}{x^{15}}+4096 x^{12}\]\[+\frac{51963120}{x^{12}}-73728 x^9-\frac{55427328}{x^9}+608256 x^6+\frac{43110144}{x^6}\]\[-3041280 x^3-\frac{24634368}{x^3}+10264320\]
:-)

Answer 41

and now I have met my quota binomial expansion problems for the month :-)