Factor x^2 + x - (a^2 - a), treating a as a constant.

Step-by-step solution

1. 1

   Clear the outer bracket carefully. The minus sign distributes over both terms inside:

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

   Missing the sign change on the $-a$ (and writing $-a$ instead of $+a$) is the main trap.

2. 2

   Treat it as a monic quadratic in x. The coefficient of $x$ is $1$ and the constant term is $a-a^2$. So we need two expressions whose sum is $1$ and whose product is $a-a^2=a(1-a)$.

3. 3

   Spot the pair. The two obvious candidates are $a$ and $1-a$:

   $a+(1-a)=1\ \checkmark,\qquad a(1-a)=a-a^2\ \checkmark$

   Both conditions hold identically in $a$, which is what makes this work for every value of the parameter.

4. 4

   Write the factorisation.

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

5. 5

   Verify by expanding symbolically.

   $(x+a)(x-a+1)=x^2-ax+x+ax-a^2+a=x^2+x-a^2+a\ \checkmark$

   The $\pm ax$ terms cancel exactly, which is why only $x$ survives in the middle.

6. 6

   Read off the roots and a useful check. The expression vanishes at $x=-a$ and $x=a-1$. Their sum is $-1=-b/a$ ✓ and their product is $-a(a-1)=a-a^2$ ✓, matching the constant term. Numerically, comparing the original with the factored form at $60$ random pairs $(x,a)$ in $[-6,6]$ gives agreement to machine precision throughout ✓.