Question

Correct answer for a medal.
What is the third term in the binomial expansion of (3x + y3)4?

54x^2*y^3
18x^2*y^3
18x^2*y^6
54x^2*y^6

Answer 1

I keep getting getting 18x^2*y^6 but it says that my answer is wrong.

Answer 2

can you show me your work?

Answer 3

I can't fix it without seeing where exactly you are going wrong

Answer 4

I think I have been writing the formula incorrectly. :/

Answer 5

ok well, tell me what you got. I need to see it, unfortunately I am not a telepath

Answer 6

@FibonacciChick666
4C2 a^2 b^3
but I don't know how to solve for C
Sorry for the wait.

Answer 7

ohk yea wrong formula, try this: \[(x+y)^4 \;=\; x^4 \,+\, 4 x^3y \,+\, 6 x^2 y^2 \,+\, 4 x y^3 \,+\, y^4.\]

Answer 8

The 4 and the 2 next to the C are supposed to be subscripts.

Answer 9

that is the binomial theorem

Answer 10

so which is the third term?

Answer 11

6x^2 y^2

Answer 12

good, so now the 'x' and 'y' are actually loose terms here they mean whatever is in the first slot and whatever is in the second slot. so what are our 'x' and 'y'?

Answer 13

2 and 2?

Answer 14

where did you get the 2's?

Answer 15

x^2 and y^2 from the third term

Answer 16

ok so let me rephrase, instead of \[(x+y)^4 \;=\; x^4 \,+\, 4 x^3y \,+\, 6 x^2 y^2 \,+\, 4 x y^3 \,+\, y^4.\] let's say\[(a+b)^4 \;=\; a^4 \,+\, 4 a^3b \,+\, 6 a^2 b^2 \,+\, 4 a b^3 \,+\, b^4.\]

Answer 17

so we have \[(3x+y^3)^4\] What is our a, and what is our b?

Answer 18

I'm sorry, but I am confused...
1 and 3?

Answer 19

it's ok so let's look at this one (5x+4y) in this case our a=5x and our b=4y

Answer 20

do you understand what I mean now?

Answer 21

oh ok so a=3x and b=y^3 right?

Answer 22

yep now just plug it in

Answer 23

6 * 3x^2 * y^6

Answer 24

not quite remember parentheses

Answer 25

6*3x^4*y^8