Question

How can you factor `a^4 + b^4` and how can you apply it to the following examples ?

`x^4+1`

`x^4+81`

`x^4+256`

Answer 1

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

Explanation:

Notice that:

`(a^2-kab+b^2)(a^2+kab+b^2) = a^4+(2-k^2)a^2b^2+b^4`

So if we put `k = sqrt(2)` then we find:

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

Hence we find:

`x^4+1 = (x^2-sqrt(2)x+1)(x^2+sqrt(2)x+1)" "` with `a=x` and `b=1`

`x^4+81 = (x^2-3sqrt(2)x+9)(x^2+3sqrt(2)x+9)" "` with `a=x` and `b=3`

`x^4+256 = (x^2-4sqrt(2)x+16)(x^2+4sqrt(2)x+16)" "` with `a=x` and `b=4`

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Historical note

No less a mathematician than the great Leibniz thought that it was not possible to factor `a^4+b^4` as a product of quadratics with real coefficients. So if you experienced any difficulty finding such a factorisation then you are in very good company.

Answer 2

This is no different, really, than the method already posted, but this is another way of thinking about the factorization process.

Note that `(a^2+b^2)^2=a^4+2a^2b^2+b^4`. We can then manipulate `a^4+b^4` as such:

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

`=(a^2+b^2)^2-(sqrt2ab)^2`

Which is a factorable difference of squares:

`=((a^2+b^2)+sqrt2ab)((a^2+b^2)-sqrt2ab)`

`=(a^2+sqrt2ab+b^2)(a^2-sqrt2ab+b^2)`