ok the first two. 3x goes into 12x^3 4x^2 times. 4x^2*(3x-5y)=12x^3-20x^2y

_4x^2-3xy+1______________

3x-5y)12x^3-11x^2y-12xy^2-5y^3

-(12x^3-20x^2y)

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0+9x^2y-12xy^2

-(9x^2y-15xy^2)

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0 + 3xy^2-5y^3

-(3xy^2-5y^3)

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0

Synthetic division works like this:

. x^5-y^5 / x-y, think of this as 1*x^5+0*x^4+0*x^3+0*x^2+0*x^1+y^5*x^0/(1*x^1-y)

Ok this works when dividing by (x-a), in your case a=y

write "a" in a box on the left and all the coefficients of the polynomial on the right:

y]1 0 0 0 0 y^5

y -y^2 y^3 -y^4 y^5

-----------------------------------------1 -y y^2 -y^3 y^4 0

so, we "drop" the 1 and multiply by y. that goes in the next space where we subtract from zero and get -y^2 which we subtract from zero getting y^2. We continue in this fashion until we run out of terms.

We now use the numbers in the bottom to form a new polynomial of one less degree. (1)*x^4+(-y)*x^3+(y^2)*x^2+(-y^3)*x^1+(y^4)*x^0+0/(x-y)

That zero over x-y comes from the remainder term. If this hadn't divided evenly the last difference in the division would be nonzero and would get placed over x-y where that zero is. Our answer, though, is

x^4-y*x^3+y^2*x^2-y^3*x+y^4.

I must mention that it is critical that for any "missing powers" in your polynomial you put zeros in your synthetic division or else it WON'T WORK.

Hope this helps.