does anyone know how to do this?
20x^3y^6-14x^5y^3+18x^4y^5/2x (divided by)
2x^3y^2

Question

supie · Answer

@Angle o:

Angle · Answer

$\huge{\frac{20x^3y^6-14x^5y^3+18x^4y^{5/2}x}{2x^3y^2}}$

Angle · Answer

a problem like this works best if you think of it as (distributive property but backwards)

for example if I factored out 2 x^3 y^2 out of just the 20 x^3 y^6
I can make it look like
20 x^3 y^6 = (2 x^3 y^2) (10 y^4)

Angle · Answer

Then
-14 x^5 y^3 can be split into (2 x^3 y^2) (-7 x^2 y)

Angle · Answer

and
18 x^4 y^5 can be split into (2 x^3 y^5) (9 x^(-1) y^3)
but I might have misinterpreted the problem there

Angle · Answer

if you are still suck, just let me know

Monicasayshi · Answer

@angle wrote:

and 18 x^4 y^5 can be split into (2 x^3 y^5) (9 x^(-1) y^3) but I might have misinterpreted the problem there

so what would the answer be? 0_o

Angle · Answer

20 x^3 y^6 = (2 x^3 y^2) (10 y^4)

-14 x^5 y^3 can be split into (2 x^3 y^2) (-7 x^2 y)

That means the first two chunks of your answer will be this stuff ^ without the (2 x^3 y^2) parts