We were discussing factoring in my Intermediate Algebra class the other day and happened upon the polynomial x4-81.
So, we factored the original expression into (x²+9)(x²-9) and then to (x²+9)(x+3)(x-3).
One student asked why we split up the x²-9, but not the x²+9, so we talked about how the x²+9 was prime whereas the x²-9 was factorable as the difference of two squares, just as we had factored the original x4-81 as a difference of two squares.
The student then went on to say, “So that means that x4+81 would be prime, just like the x²+9 was prime….”
I hesitated because I remembered seeing a textbook exercise somewhere (and I still can’t remember where) which factored a sum of x4 and a constant of some kind.
So, I mentioned this, said that we shouldn’t overgeneralize and promised to look into it.
After class, I went to Wolfram Alpha and typed in “factor x^4+81” and got back (among other things) the factorization of
or, if we look at this in general terms,
(x²-bx+c)(x²+bx+c) where 2c=b²
Next, I worked to set this up so that b would be a whole number, in particular b=2 and c=2. This leads to an expression of
I then remembered the factorization I had seen in the textbook was as follows:
take x4+4 and add and subtract 4x²
then factor the first three terms
then factor as a difference of squares
Generalizing this a little bit, I came up with a form of x4+4n4 factorable as (x²+2xn+2n²)(x²-2xn+2n²) by adding and subtracting 4x²n².
This generaliztion comes from the relationship between “b” and “c.” If 2c=b², then for both “b” and “c” to be whole numbers, b² must be even, and therefore must be an even perfect square (4, 16, 36, 64, ….).
If b² is even, then “b” itself must also be even…so
And, because the final term in each expression (x4+81 or x4+4) comes from squaring “c”
So, in general