Synthetic Division for (x^4 - y^4) / (x - y)

[axrw]

I am supposed to use synthetic division to find the quotient and remainder.

x^4 - y^4 / x - y

I haven't done this with two variables before, I'm not really sure how to go about it.

Thanks for any help.

 

[tkhunny]

Identify the coefficients.

x^4 + 0x^3y + 0x^2y^2 + 0xy^3 - y^4

Use 'y'

 

[axrw]

I really appreciate your help.

I think I'm just slow, so please be patient with me, but I'm still not getting this.

y | 1          0          0          0          -1

               y          y^2      y^3       y^4
_____________________________________________________
     1          y          y^2      y^3      | y^4 - 1

That result doesn't really make sense to me.

I did the long division instead and got:

x^3 + yx^2 + xy^2 + y^3

with no remainder.

But it's really bugging me I can't understand this.

 

[stapel]

I'm not really sure how you're supposed to use synethetic division for this, either, frankly. :shock: Synthetic division allows you to divide (or evaluate) a one-variable polynomial by linear factor (or at a real-valued number, respectively).

At a guess, the "one" variable is the "x", and the "y" is meant to be dealt with as though it were a number, like 2, 2², 2³, etc. So you'd be dividing by x - y (or evaluating at x = y), and dividing into x⁴ + 0x³ + 0x² + 0x - y⁴, with "y⁴" being, in effect, the constant term. If this is correct, then you'd have the following:

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

Recalling that this last line here displays only the constants of the cubic (in x) that is the quotient, the division would then yield x³ + x²y + xy² + y³, with a zero remainder. Since this matches what you got "the long way", this is probably valid.

...still seems an odd way to use synthetic division, though....

Eliz.

 

[axrw]

Thank you stapel and tkhunny. It makes more sense now.