Factor x4 + x2 - 2ax + 1 - a2 completely.

Step-by-step solution

1. 1

   Look for a square to complete, guided by the end terms. The $x^4$ and the $+1$ suggest the square $\left(x^2+1\right)^2=x^4+2x^2+1$. Compare it with what we have: the given expression has only $x^2$, so it is short by $x^2$, and it carries the extra $-2ax-a^2$.

2. 2

   Do the bookkeeping explicitly.

   $x^4+x^2-2ax+1-a^2=\left(x^4+2x^2+1\right)-x^2-2ax-a^2=\left(x^2+1\right)^2-\left(x^2+2ax+a^2\right)$

   and the bracket on the right is itself a perfect square, $(x+a)^2$. So the whole thing is

   $\left(x^2+1\right)^2-(x+a)^2$

   This is the key step, and it shows the claim that the expression "does not factor nicely for arbitrary $a$" is false.

3. 3

   Apply the difference of squares. With $A=x^2+1$ and $B=x+a$:

   $A^2-B^2=(A-B)(A+B)=\left(x^2+1-x-a\right)\left(x^2+1+x+a\right)$

4. 4

   Tidy each bracket.

   $\left(x^2-x+1-a\right)\left(x^2+x+1+a\right)$

   Both factors are quadratics in $x$ with the parameter sitting only in the constant term.

5. 5

   Verify by expanding. Multiplying out, the $x^3$ terms cancel ($-x^3+x^3$), the $x^2$ terms give $x^2+x^2-x^2=x^2$ after the cross terms, the $x$ terms collapse to $-2ax$, and the constants give $(1-a)(1+a)=1-a^2$ — reproducing the original exactly. A random numerical check ($x=1.5$, $a=0.7$) gives the same value on both sides.

6. 6

   Sanity-check the special case $a=0$. The factorization becomes

   $\left(x^2-x+1\right)\left(x^2+x+1\right)=x^4+x^2+1$

   which is the well-known identity — and indeed setting $a=0$ in the original also gives $x^4+x^2+1$ ✓.