Question

How do you long divide `(a^4 + 4b^4) / (a^2 - 2ab + 2b^2)`?

Answer 1

Long divide coefficients to find:

`(a^4+4b^4) / (a^2-2ab+2b^2) = a^2+2ab+2b^2`

Explanation:

Notice that the numerator and denominator are both homogeneous polynomials: The degree of the terms of the numerator are all `4` and of the denominator are all `2`. So we can just look at the coefficients and long divide them, not forgetting to add `0`'s for the "missing" terms in the denominator (for `a^3b`, `a^2b^2` and `ab^3`) ...

The process is similar to long division of numbers:

Write the dividend under the bar and the divisor to the left. The dividend `1, 0, 0, 0, 4` represents `a^4+0a^3b+0a^2b^2+0ab^3+4b^4`. The divisor `1, -2, 2` represents `a^2-2ab+2b^2`.

Write the first term `1` of the quotient above the bar. Choosing it so that when multiplied by the divisor it matches the first term of the dividend.

Write the product (`1, -2, 2`) of this initial term and the divisor under the dividend and subtract it.

Bring down the next term `0` from the dividend alongside it.

Choose the next term of the quotient to match the leading term of the running remainder, etc.

Eventually we run out of terms in the dividend to bring down. The last subtraction is exact, so there is no remainder and the division is exact.

We find:

`(a^4+4b^4) / (a^2-2ab+2b^2) = a^2+2ab+2b^2`

Footnote

This particular example is interesting in showing the factorisation:

`a^4+4b^4 = (a^2-2ab+2b^2)(a^2+2ab+2b^2)`

In general we find that:

`(x^2-kxy+y^2)(x^2+kxy+y^2) = x^4+(2-k^2)x^2y^2+y^4`

So if you are faced with a quartic polynomial that has no odd terms and want to factor it, remember this identity.

In particular, putting `k = sqrt(2)` eliminates the middle term too.

In our example we have a case with `x=a`, `y=sqrt(2)b`, `k=sqrt(2)`