Question

using the expanding binomial theorem what is (3v-2u)^4

Answer 1

do you know the "expansion" formula (from your notes) ?

Answer 2

yes but not for using 2 variables

Answer 3

the general form is
nChooseK (a)^(n-k) (b)^k

where k goes from 0 to n

to use it, match up a with 3v
and b with (-2u)

Answer 4

how do i simplify it

Answer 5

n is 4

so the first term (with k=0) is
4C0 * (3v)^(4-0) * (-2u)^0

4C0 is 1 (do you know how to do this part ? )
(3v)^4 means (3v) * 3v *3v*3v or 3^4 v^4
using a calculator or multiplying out , 3*3*3*3= 9*9= 81
we get

1 * 81*v^4 * (-2u)^0
anything to the zero power is 1, so the first term is
81 v^4

Answer 6

do you follow how to do the first term ?

Answer 7

ok so i just keep doing that to the rest of the terms right

Answer 8

Let's do k=1:
4C1 * (3v)^(4-1) * (-2u)^1

can you simplify each part ?

Answer 9

what is 4C1 ?

Answer 10

4C1= (3v)^3 right

Answer 11

no
4C1 or 4 choose 1 or
\[\left(\begin{matrix}4 \\ 1\end{matrix}\right)= \frac{ 4! }{ 1! (4-1)! }\]

Answer 12

or use your calculator (look for the key labeled nCr, for example)

Answer 13

what is 4C1 ?

Answer 14

i get 216v^2u^2 when i simplify the term

Answer 15

what did you get for 4C1 ?

Answer 16

4[(3v)^3 (-2)^1]

Answer 17

4* 3^3 *(-1)^1 is what ?

Answer 18

108

Answer 19

I can't type today.
4* 3^3 *(-2)^1 is what ?

Answer 20

-108 is what i got

Answer 21

4*3*3*3*-2

Answer 22

-216

Answer 23

now we have
4C1 * (3v)^(4-1) * (-2u)^1
(4*3^3*-2) * v^3 * u^1
or
-216 * v^3 * u^1

we leave out the multiply signs (the *'s)
-216v^3 u
(we leave out the exponent if it's 1... or we could put it in : -216v^3 u^1

Answer 24

can you write down the formula when k=2 ?

Answer 25

6[(3v)^2 (-2u)^2] which goes to 6*3*3*-2*-2

Answer 26

yes, and for the letters?

Answer 27

which is positive 216v^2u^2

Answer 28

yes
so far we have
\[ 81v^4 -216v^3 u +216 v^2 u^2 + ...\]
now k= 3

Answer 29

the "letter part" follows a pattern.. we can write them down without thinking
the V's exponent goes down by 1, the U's exponent goes up by 1... until we get to V^0 (which is 1)

Answer 30

4[(3v)^1 (-2u)^3] which is 4*3*-2*-2*-2= -96vu^3 like that?

Answer 31

yes
\[ 81v^4 -216v^3 u +216 v^2 u^2 -96 v u^3+\]
one last term k=4

Answer 32

1[(3v)^0 (-2u)^4] which is 1*3^0*-2*-2*-2*-2= positive 16u^4

Answer 33

and the whole expansion is
\[ 81v^4 -216v^3 u +216 v^2 u^2 -96 v u^3+ 16u^4 \]

Answer 34

the letters are obviously correct: v goes down from exponent 4 to 0 (last term v^0 is 1)
and u goes up from exponent 0 to 4
the signs alternate (because -2^ odd number is negative and -2 to even exponent is +)
so if we did not make a mistake with the numbers (I don't think we did), it looks good.

Answer 35

ok thank you.. you taught me better than my teacher did