how to factor $f(x)=x^4-7x^2+1$

Question

How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.

Answer 1

It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+\sigma)(x^2-bx+\sigma)$ with $\sigma=\pm1$. We have

$$(x^2+bx+\sigma)(x^2-bx+\sigma)=x^4+(2\sigma-b^2)x^2+1$$

so we need $2\sigma-b^2=-7$, which rewrites as $b^2=7+2\sigma$. Letting $\sigma=1$ gives $b=\pm3$, so we have the factorization

$$x^4-7x^2+1=(x^2+3x+1)(x^2-3x+1)$$

Note, the other option, $\sigma=-1$, gives another valid factorization, just not over the integers:

$$x^4-7x^2+1=(x^2+\sqrt5x-1)(x^2-\sqrt5x-1)$$