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