<Question>
Factor the expression $ab+5b+2a+10$.
</Question>

<think>
<Plan_1>
Okay, so I need to factor the expression ab + 5b + 2a + 10. Hmm, let's see. Factoring can sometimes be tricky, but maybe I can group terms together or find common factors. Let me start by looking at each term and seeing if there's a common factor in pairs.

First, let's list out the terms: ab, 5b, 2a, and 10. I notice that the first two terms have a 'b' in them. Maybe I can factor out a 'b' from those?
Let's try that: ab + 5b can be written as b(a + 5). Then, the last two terms are 2a and 10. Those have a '2' in common. If I factor out a 2 from those, that would be 2(a + 5). Wait, so then the expression becomes b(a + 5) + 2(a + 5).

Hmm, now I see that both terms have an (a + 5) factor. So if I factor out (a + 5), then the remaining terms would be b and 2. So that would give (a + 5)(b + 2).

Let me check to make sure I did that right.
Let's expand (a + 5)(b + 2) to see if we get back the original expression. Multiplying a by b gives ab, a by 2 is 2a, 5 by b is 5b, and 5 by 2 is 10. So combining those terms: ab + 2a + 5b + 10. Which is the same as the original expression, just written in a different order. So yes, that factors correctly.

But let me go through the steps again to ensure I didn't skip anything. Starting with ab + 5b + 2a + 10. Group the first two terms and the last two terms: (ab + 5b) + (2a + 10). Factor out the common factors from each group.

<prompt_1.1> Re-evaluate the factoring process by grouping the expression and factoring out common factors again. </prompt_1.1>

Another way to approach this might be to rearrange the terms to see if that helps. Let's try grouping ab + 2a first and then 5b + 10. So ab + 2a can be factored by taking out an 'a', giving a(b + 2). Then 5b + 10 can have a 5 factored out, giving 5(b + 2). So now we have a(b + 2) + 5(b + 2), which again factors to (a + 5)(b + 2).
-----

Is there any other way this could be factored? Let me think. Suppose I tried a different grouping. For example, if I took ab + 10 together and 5b + 2a. But ab + 10 doesn't have an obvious common factor. Let's see: ab + 10. The factors of 10 are 1 and 10, 2 and 5. But unless a is 2 or 5, that might not help. Similarly, 5b + 2a. Maybe factor out a 'b' from the first group and something else from the second? Not sure.
Let's try: If I group ab + 10 and 5b + 2a. For ab + 10, maybe factor out a, giving a(b) + 10. Not helpful. For 5b + 2a, maybe factor out a 5? 5(b) + 2a. Still not obvious. If I try to factor these, it might not lead anywhere.

Alternatively, maybe factor out a 'b' from all terms. Let's see: ab +5b is b(a +5). Then 2a +10 is 2(a +5). Wait, that's the same as before. So regardless of how I group, it seems like (a +5) and (b +2) are the common factors.
</Plan_1>

<Execution_1>
<execution_1.1>
From ab + 5b, factor out a 'b', which gives b(a + 5). From 2a + 10, factor out a 2, which gives 2(a + 5). So now the expression is b(a + 5) + 2(a + 5). Then, since both terms have (a + 5), factor that out: (a + 5)(b + 2).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me verify once more by expanding.
(a + 5)(b + 2) = a*b + a*2 + 5*b + 5*2 = ab + 2a + 5b + 10. Yep, that's exactly the original expression.
-----

I don't think there's any need to complicate it further. Sometimes, expressions can have multiple factoring methods, but in this case, both groupings lead to the same result.
So I think that's the correct answer.
</Plan_2>

<Execution_2>
No prompt to execute.
</Execution_2>

<Final_answer>
ab + 5b + 2a + 10 = \boxed{(a+5)(b+2)}.

\]
</Final_answer>
</think>