Question

Expand the binomial (2x + y2)5.

Answer 1

From the binomial theorem:
\[ (2x+y^2)^5=\sum_{r=0}^5{5\choose r}(2x)^{5-r}(y^2)^r\\
~ \\ = {5\choose 0}(2x)^5(y^2)^0+{5\choose 1}(2x)^4(y^2)^1+\ldots+{5\choose 5}(2x)^0(y^2)^5\]

Answer 2

I'm so incredibly confused. i don't know how you got that. would i solve part of that or is the answer in there? i'm sorry! i'm trying...not very smart.

Answer 3

Have you seen the binomial theorem: Which says to find the expansion of something of the form: \[(x+y)^n=\sum_{r=0}^n {n \choose r}x^{n-r}y^r\]

Answer 4

If the formula doesn't look familiar at all, then don't worry. We can solve your problem another way.

Answer 5

It doesn't look familiar. i'm sorry.

Answer 6

So what you can do is just "manually" expand your polynomial. Do you agree with the following:
\((2x+y^2)^5=(2x+y^2)^2(2x+y^2)^2(2x+y^2)^1\)

(This is just from the law of exponents \(a^xa^y=a^{x+y}\), Here pretend that \(a=2x+y^2\), so in fact I decomposed your polynomial in this way: \(a^5=a^{2+2+1}=a^2a^2a^1\)

Answer 7

this makes much more sense the one before but i may still need help solving it...i apologize

Answer 8

it's asking me what the fourth term would be...i don't get it

Answer 9

Don't worry, it's possible that you haven't seen that formula yet :) It's just a formula that more quickly expands your polynomial.

But yeah, if you agree with that, then what you can do is expand \((\color{red}{2x}+\color{blue}{y^2})^2=\color{red}{(2x)}^2+2\color{red}{(2x)}\color{blue}{(y^2)}+\color{blue}{(y^2)}^2\)
This is simply from:
\((\color{red}{a}+\color{blue}{b})^2=\color{red}{a}^2+2\color{red}{a}\color{blue}{b}+\color{blue}{b}^2\)

And then, you can multiply each of your factors using the distributive property of multiplication.

Answer 10

i am trying that right now:) i'm very behind in school and doing my best to get caught up. thank you for your patience with me. one moment while i solve:)

Answer 11

so everything in blue you can combine as well as red?

Answer 12

I'm not exactly sure what you mean :S

But I just used the colour code so that you could see how to use the formula for \((a+b)^2\) in your case , where \(a=2x\) and \(b=y^2\)

Answer 13

okay well ill let you know what i get and can you let me know if it's right?

Answer 14

ok

Answer 15

i'm getting frustrated. lol i am gonna go on to a different subject for a bit. thanks so much for ur help

Answer 16

aw ok . well glad I could help somehow :P