Factor x2 - 2ax + x + a2 - a - 2 completely.

Step-by-step solution

1. 1

   Collect the terms by powers of $x$.

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

   It is a quadratic in $x$ whose coefficients involve $a$ — so any factorization should be a product of two linear-in-$x$ brackets.

2. 2

   Spot the completed square hiding inside. The pieces $x^2$, $-2ax$ and $a^2$ are exactly $(x-a)^2$. Peeling them off leaves

   $x^2-2ax+a^2+x-a-2=(x-a)^2+(x-a)-2$

   because the leftover $+x-a$ is itself $(x-a)$. Every occurrence of $x$ and $a$ now sits inside the single combination $x-a$.

3. 3

   Substitute $u=x-a$ and factor the plain quadratic.

   $u^2+u-2=(u-1)(u+2)$

   since $(-1)(2)=-2$ and $-1+2=1$. With the substitution the parameter has disappeared entirely.

4. 4

   Back-substitute $u=x-a$.

   $(x-a-1)(x-a+2)$

5. 5

   Cross-check two ways. By the coefficient route: $a^2-a-2=(a-2)(a+1)$, and the two bracket constants $-(a+1)$ and $2-a$ sum to $1-2a$ ✓ matching the linear coefficient. Numerically at $x=1$, $a=2$: the original is $1-4+1+4-2-2=-2$, and $(1-2-1)(1-2+2)=(-2)(1)=-2$ ✓.