This polynomial is in quadratic form. If we use the substitution `t=x^2` we get `t^2+8t-9` .
Factor this using any methods you like: e.g. find numbers p,q such that pq=-9 and p+q=8. Then p=9 and q=-1 so:
`t^2+8t-9=(t+9)(t-1)` Substituting for t we get:
`(x^2+9)(x^2-1)` The second binomialis the difference of two squares:
`(x^2+9)(x+1)(x-1)` If we are in the complex field we can factor the sum of two squares:
So `x^4+8x^2-9=(x^2+9)(x+1)(x-1)` in the reals.
This works for any polynomial with real coefficients that is factorable over the rational numbers that is in quadratic form.
For example `x^6+8x^3-9` is quadratic in `x^3` : if `t=x^3` then again we have `t^2+8t-9=(t+9)(t-1)` so `x^6+8x^3-9=(x^3+9)(x^3-1)` etc...
If you have polynomial
`P(x)=a_nx^n+a_(n-1)x^(n-1)+cdots+a_1x+a_0`, `a_n,ldots,a_0 in CC`
you can factor it by finding solutions of equation `P(x)=0`.
If `x_1,x_2,ldots,x_n` are solutions of equation `P(x)=0`, then you can write your polynomial as `P(x)=(x-x_1)(x-x_2)cdots(x-x_n)`. But finding exact solutions to equation `P(x)=0` isn't always easy especially for polynomials of degree 5 and higher because there are no formulas for solution of such equations (it' not that formula hasn't been found, but it doesn't exist that is it can never be found).
In your case we are looking for solutions of equation
let's make substitution `t=x^2` to get quadratic equation
`t_1=-9 => x_(1,2)=pm sqrt(-9)=pm3i`
`t_2=1 => x_(3,4)=pm sqrt(1)=pm1`
Now that we have all four solutions we can factor our polynomial