1. ## Factor *!URGENT!*

i need help with this problem

Factor:
(3ax^2+2b^2y+axy)^2

2. Originally Posted by AHDDM
i need help with this problem

Factor:
(3ax^2+2b^2y+axy)^2
Not sure what you want.
$\displaystyle [ax(3x+y)+2b^2y]^2$
$\displaystyle (ax)^2(3x+y)^2+4ab^2xy(3x+y)+4b^4y^2$

3. Hello, AHDDM!

Are you sure the instructions say "factor"?

Factor: .$\displaystyle (3ax^2 + 2b^2y + axy)^2$

Simple!

$\displaystyle (3ax^2 + 2b^2y + axy)^2\;=\;(3ax^2 + 3b^2y + axy)$$\displaystyle \cdot(3ax^2 + 2b^2y + axy)$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

It's more likely that they want us to multiply . . .

After a lot of multiplying and combining, we get:
. . $\displaystyle 9a^2x^4 + 12ab^2x^2y + 6a^2x^3y + 4b^4y^2 + 4ab^2xy^2 + a^2x^2y^2$

4. i got wat soroban got after multipling so ima go with tat.

can you guys also help me with this

$\displaystyle (2x-7)^6$

5. Originally Posted by AHDDM
$\displaystyle (2x-7)^6$
what's the question, that number is already factored...

6. if its already factored then im ok thats what i got from $\displaystyle 64x^6-y^6$

7. Originally Posted by AHDDM
if its already factored then im ok thats what i got from $\displaystyle 64x^6-y^6$
No that is wrong.
You need to use the binomial theorem.

8. Originally Posted by Quick
what's the question, that number is already factored...
I'm sorry, let me rephrase this to be:
Originally Posted by Quick
what's the question, that number is already as simple as it will get...