Question

Factorize completely

a^2 - ax - 1 + x

Answer 1

To factor the expression \( a^2 - ax - 1 + x \), we can start by rearranging it for better clarity:

\[
a^2 - ax + x - 1.
\]

Next, we can group the terms to facilitate factoring:

\[
a^2 - ax - 1 + x = a^2 - ax + (x - 1).
\]

Now, let's rearrange the terms a bit differently:

\[
a^2 - ax + x - 1 = a^2 - ax + x - 1.
\]

At this point, we can try factoring by recognizing a common factor from the \(a\) terms:

1. Group the first two terms together and the last two terms together:
\[
(a^2 - ax) + (x - 1).
\]

2. Factor out \(a\) from the first group and \(1\) from the second group:
\[
a(a - x) + 1(x - 1).
\]

Now, observe that we can rearrange \( (x - 1) \) to help us factor:
\[
x - 1 = -(1 - x).
\]

So we have:
\[
a(a - x) - (1 - x).
\]

If we look for a common form:
Notice if we can get a shared factor:

This is now structured but does not lead to a simple second factorization. Let's try pairing them differently.

If we manipulate this, we can see:

Expanding back:
\[ a(a - x) + (x - 1). \]

Now separating \(x\) gives us:

\[
a^2 - ax + x - 1 = a^2 + (1 - a)x - 1.
\]

Attempting to find values for common roots will reveal that \(x - 1\) is a factor.

More directly:
*Consider adding completing the square or synthetic division methods to check for factors.*

1. If we apply quadratic formula explicitly:

\[
a^2 - (x + 1)a - 1 = 0 \to a = \frac{(x + 1) \pm \sqrt{(x + 1)^2 + 4}}{2}.
\]

Would yield roots, with symmetric terms.

Ultimately the algebra brings out to :

Final factorization:
\[
(a - (x + 1))(a + (x - 1)) \text{ with roots measured.}
\]

Thus, the complete factorization requires awareness of linear terms, completing analysis to help factorize completely if denoting directly to numerical roots.

So, the expression can be succinctly inferred as a result leading towards simple formulative arrangements generally yielding fragmants as seen:

\[
(a - (x + 1))(a + (x - 1)).
\]