Use synthetic division to find the quotient and remainder: $\displaystyle (x^4-y^4)/(x-y)$

I know how to do synthetic division with just x, but this threw me for a loop. Bagatrix (math software) also could not solve it. Any ideas?

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- May 2nd 2010, 10:53 AMChristopherDunnSynthetic division in two variables?
Use synthetic division to find the quotient and remainder: $\displaystyle (x^4-y^4)/(x-y)$

I know how to do synthetic division with just x, but this threw me for a loop. Bagatrix (math software) also could not solve it. Any ideas? - May 2nd 2010, 11:41 AMstapel
My

*guess*(and it's only a guess!) is that you are expected to use the underlying concepts for synthetic division, and apply them here.

When you do**synthetic division**of a polynomial in x by a factor in x, you are actually just plugging in the corresponding x-value. For instance, if you divide by x - 3, you are actually only plugging the 3 into the division and, if the remainder is zero, then you will have shown that x = 3 is a zero of the original polynomial (or, which is the same thing, that x - 3 is a factor).

In this case, you are dividing by x - y. When you were dividing by x - 3, you set this equal to zero, and solved for the value to plug into the synthetic-division matrix: x - 3 = 0, so x = 3, so plug in "3". The corresponding process here would be: x - y = 0, so x - y, so plug in "y".

You can find "the answer" by long polynomial division. Try the synthetic division with the top row being 1, 0y, 0y^2, 0y^3, 1y^4, and see if you get the right result. (Wink) - May 2nd 2010, 12:05 PMChristopherDunn
Aha! Right you are (with the exception of the sign on the last coefficient). I used $\displaystyle y$ for my synthetic divisor and used $\displaystyle [1][0][0][0][-y^4]$ for my coefficients. Carrying it through I got $\displaystyle x^3+x^2y+xy^2+y^3$ with no remainder. Thanks!