Question

really need help! simply: (3m^3-1/2y)(3m^3-1/2y)

Answer 1

*simplify, my bad

Answer 2

We have
\[(3m^3-\frac{1}{2} y)(3m^3-\frac{1}{2} y)\]
Do you know the expansion of (a-b)^2?

Answer 3

nooo? What is that?

Answer 4

It's a standard relation always remember it
\[(a-b)^2=(a-b)(a-b)=a^2-ab-ba+b^2=>a^2-2ab+b^2\]

Answer 5

oh gosh

Answer 6

We'll use that here
\[(3m^2-\frac{1}{2}y)(3m^2-\frac{1}{2}y)={(3m^2-\frac{1}{2}y)}^2\]
Did you not understand it?

Answer 7

@becca94 ???

Answer 8

sorry I was helping someone else really quick. umm... I kind of understand it. I'm not good at AlgII

Answer 9

I've never heard of the standard relation and that's a lot in a specific order to remember.

Answer 10

Ok
\[(3m^2-\frac{1}{2}y)^2\]
here
\[a=3m^2\]
\[b=-\frac{1}{2} y\]

so using the formula, we get
\[(3m^2-\frac{1}{2}y)^2= 9m^4-3m^2y+\frac{1}{4}y^2\]

Answer 11

okay...I'm understand a bit better now

Answer 12

Just remember
\[(a-b)^2=a^2-2ab+b^2\]

Answer 13

I'll write it down!

Answer 14

Great:D