Question

Find the constant term in the expansion of (2x^2 - 1/x)^6

Answer 1

Ok so I know that you're supposed to use this equation:

Tsub r+1 = \[\left(\begin{matrix}n \\ r\end{matrix}\right)\]\[a^{n-r}b^{r}\]

Answer 2

I dunno why that came out so weird. Anyway, I plugged in a, so 2x^2, b, and n, but my answer is completely wrong. It's 60 but I got r=6

Answer 3

yes, and your job is to figure out what \(r\) is

Answer 4

yea but I did it wrong. I tried to find the general term but..I don't think my algebra is right

Answer 5

\(n=6\) in your example

Answer 6

yes, n=6 in the example

Answer 7

and what you are looking for is where the exponent on the \(\frac{1}{x}\) term is equal to the exponent on the \(x^2\) term so that you will get 1 when you divide

Answer 8

umm yes, I guess so

Answer 9

sure because you will have a term that looks like \[C\frac{x^m}{x^m}\] which is just \(C\)

Answer 10

I've never seen that equation before but ok

Answer 11

we can do it using algebra, or we can just guess what it has to be

Answer 12

it is not an equation, i am just saying if the exponent is the same top and bottom then the x terms will cancel and we will have a constant

Answer 13

no lets use algebra lol

Answer 14

ok cool

Answer 15

so the first term is in the parentheses is \(2x^2\) meaning you will have \[(2x^2)^{6-r}\] or \[2^{6-r}x^{12-2r}\] for each of those
clear or no?

Answer 16

yup

Answer 17

i know you distribute but do you get 64x^12-2r?

Answer 18

if you write \(\frac{1}{x}\) as \(x^{-1}\) then those terms will look like \(x^{r}\)

Answer 19

no it is not 64 because we don't know what r is

Answer 20

wait I'm confused. so you don't distribute the 6 to the 2?

Answer 21

my book shows that even if you don't know what r is you still distribute it

Answer 22

ok 2 is being raised to the power of \(6-r\)

Answer 23

wait so does it just stay \[2x^{12-2r}\]

Answer 24

nope you have to raise 2 to the power of \(6-r\) as well

Answer 25

right so then it's 64

Answer 26

just like say \[(ab^2)^4=a^4b^8\]

Answer 27

it is not 64 since we do not know what \(r\) is yet

Answer 28

...um could you just show me what it looks like cos I'm confused

Answer 29

i did but i will be happy to write it again \[\huge (2x^2)^{6-r}=2^{6-r}(x^2)^{x-r}=2^{6-r}x^{12-2r}\]

Answer 30

ok makes sense, thank you

Answer 31

too large \[\large (2x^2)^{6-r}=2^{6-r}(x^2)^{x-r}=2^{6-r}x^{12-2r}\]

Answer 32

and the other one will be \[\large x^{-r}\]

Answer 33

when you multiply of course you add the exponents, and you want that product to be 0 because \(x^0=1\) then you will get your constant

Answer 34

wait im confused, why is x^-r?

Answer 35

why is the 6 not added?

Answer 36

or distributed in this case

Answer 37

\[\frac{1}{x}=x^{-1}\] so \[(\frac{1}{x})^r=x^{-r}\]

Answer 38

ohhhh ok thanks

Answer 39

don't forget you wrote \[a^{n-r}b^{r}\] here \(n=6\) so it is \[a^{6-r}b^r\]no 6 in the second part

Answer 40

kk

Answer 41

next job is to solve \[12-2r-r=0\] for \(r\)

Answer 42

what about the 6-r?

Answer 43

you said you wanted to do it the algebra way, so that is what we are doing

Answer 44

also where did the third -r come from in the 12-2r-r equation?

Answer 45

the \(6-r\) got turned in to \(12-2r\) when we distributed the 2 from the \(x^2\) term

Answer 46

that was this line here \[\large (2x^2)^{6-r}=2^{6-r}(x^2)^{x-r}=2^{6-r}x^{12-2r}\]

Answer 47

ok I'm sorry I'm really confused. In the equation you gave me about, you said \[2^{6-r}\]. but then you said you dont distribute the 6 to the 2, so it therefore won't become 64, so you leave it as that. so i don't know how you're getting 6-r into 12-2r if you weren't supposed to do anything that it in the first place

Answer 48

lets go slower

Answer 49

suppose for example \(r=2\)

Answer 50

you would have \[\binom{6}{2}(2x^2)^4(-\frac{1}{x})^2\]

Answer 51

ok um, it's almost 11 where I'm at, and we've been doing this for 20 minutes, so I think I'll just come back later. sorry to have wasted your time but i need to continue. ill be back in a while but im going to close the question because this is just confusing me. thanks again