Question

How do you factor `x^8+2x^6+3x^4+2x^2+1`?

Answer 1

`x^8+2x^6+3x^4+2x^2+1 = (x^2-x+1)^2(x^2+x+1)^2`

Explanation:

Given:

`x^8+2x^6+3x^4+2x^2+1`

Note that:

• The degree of all of the terms is even, so this can be treated as a quartic in `x^2`.

• The coefficients are symmetric and in the pattern `1, 2, 3, 2, 1`.

• `12321 = 111^2`

Hence we find:

`x^8+2x^6+3x^4+2x^2+1 = (x^4+x^2+1)^2`

Note that `x^4+x^2+1 >= 1` for all real values of `x`, so has no real zeros and no linear factors with real coefficients. It can be factored into quadratics by completing the square and using the difference of squares identity:

`x^4+x^2+1 = x^4+2x^2+1-x^2`

`color(white)(x^4+x^2+1) = (x^2+1)^2-x^2`

`color(white)(x^4+x^2+1) = ((x^2+1)-x)((x^2+1)+x)`

`color(white)(x^4+x^2+1) = (x^2-x+1)(x^2+x+1)`

So:

`x^8+2x^6+3x^4+2x^2+1 = (x^2-x+1)^2(x^2+x+1)^2`

Notes

The fact that `12321 = 111^2` is directly related to:

`t^4+2t^3+3t^2+2t+1 = (t^2+t+1)^2`

as you can see by putting `t=10`

in our example, we can put `t=x^2` to find:

`x^8+2x^6+3x^4+2x^2+1 = (x^4+x^2+1)^2`

Another way of looking at this is an analogue of Pascal's triangle.

Pascal's triangle tells you the coefficients of powers of a binomial. Each number in Pascal's triangle is the sum of the two numbers above it to the left and right.

For a trinomial where the terms are in geometric progression, we can instead use a variant of Pascal's triangle in which each term is the sum of the `3` terms above it to the left, centre and right:

`color(white)(000000000)1`
`color(white)(000000)1color(white)(00)1color(white)(00)1`
`color(white)(000)color(red)(1)color(white)(00)color(red)(2)color(white)(00)color(red)(3)color(white)(00)color(red)(2)color(white)(00)color(red)(1)`
`1color(white)(00)3color(white)(00)6color(white)(00)7color(white)(00)6color(white)(00)3color(white)(00)1`