## Factoring

October 30, 2009 by richbeveridge

We were discussing factoring in my Intermediate Algebra class the other day and happened upon the polynomial x^{4}-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 x^{4}-81 as a difference of two squares.

The student then went on to say, “So that means that x^{4}+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 x^{4} 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

(x²-3(√2)x+9)(x²+3(√2)x+9)

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

x^{4}+4

factorable as

(x²+2x+2)(x²-2x+2)

I then remembered the factorization I had seen in the textbook was as follows:

take x^{4}+4 and add and subtract 4x²

x^{4}+4x²+4-4x²

then factor the first three terms

(x^{4}+4x²+4)-4x²

(x²+2)²-4x²

then factor as a difference of squares

(x²+2+2x)(x²+2-2x)

or

(x²+2x+2)(x²-2x+2)

Generalizing this a little bit, I came up with a form of x^{4}+4n^{4} 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

b=2n

b²=4n²

2c=b²=4n²

2c=4n²

c=2n²

And, because the final term in each expression (x^{4}+81 or x^{4}+4) comes from squaring “c”

c²=4n^{4}

So, in general

x^{4}+4n^{4}=x^{4}+4x²n²-4x²n²+4n^{4}^{ }

or

x^{4}+4x²n²+4n^{4}-4x²n²

(x²+2n²)²-4x²n²

(x²+2n²+2xn)(x²+2n²-2xn)

or

(x²+2xn+2n²)(x²-2xn+2n²)

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