Okay, so I need to factor the expression ab + 5b + 2a + 10. Let me start by looking at the terms and see if there's a common factor or a way to group them. Hmm, the terms are ab, 5b, 2a, and 10. Let me see if I can group them in pairs. Maybe group the first two terms and the last two terms? Let's try that.

First pair: ab + 5b. I notice that both terms have a 'b' in them. So I can factor out a 'b' from the first pair. That would give me b(a + 5). Then the second pair is 2a + 10. Both of these have a common factor of 2, so factoring that out gives 2(a + 5). Wait a minute, both pairs have (a + 5) as a factor now. That's promising. So after factoring the pairs, we have b(a + 5) + 2(a + 5). Now, since both terms have (a + 5), we can factor that out. So factoring out (a + 5) gives (a + 5)(b + 2). Let me check that by expanding it to see if it matches the original expression.

Expanding (a + 5)(b + 2): a*b + a*2 + 5*b + 5*2 = ab + 2a + 5b + 10. Wait, that's ab + 5b + 2a + 10, which is exactly the original expression. So that works! So the factored form is (a + 5)(b + 2).

But let me just make sure I didn't make any mistakes. Let me try another grouping. Suppose I group ab + 2a and 5b + 10. Then for the first group, ab + 2a, factor out an 'a' to get a(b + 2). For the second group, 5b + 10, factor out a 5 to get 5(b + 2). Then we have a(b + 2) + 5(b + 2), which again factors to (b + 2)(a + 5). Same result. So that confirms it.

Alternatively, maybe there's another way to approach this, like using the distributive property in reverse. Let's see. If I look for terms that can be grouped based on their coefficients. For example, terms with 'a' and terms without 'a'. Let's see: ab + 2a + 5b + 10. The terms with 'a' are ab and 2a. Let's factor out an 'a' from those: a(b + 2). The terms without 'a' are 5b and 10. Factor out a 5: 5(b + 2). So then we have a(b + 2) + 5(b + 2), which again is (a + 5)(b + 2). Same result.

Another way could be to use the method of factoring by grouping. The original expression is ab + 5b + 2a + 10. Let's arrange the terms as (ab + 2a) + (5b + 10). Then factor out 'a' from the first group and '5' from the second group: a(b + 2) + 5(b + 2). Then factor out (b + 2), giving (a + 5)(b + 2). Yep, same thing.

I think this is solid. All different grouping methods lead to the same factored form. Let me just check once more by plugging in some numbers. Suppose a = 1 and b = 1. Original expression: 1*1 + 5*1 + 2*1 + 10 = 1 + 5 + 2 + 10 = 18. Factored form: (1 + 5)(1 + 2) = 6*3 = 18. Same result. Another test: a = 2, b = 3. Original: 2*3 + 5*3 + 2*2 + 10 = 6 + 15 + 4 + 10 = 35. Factored: (2 + 5)(3 + 2) = 7*5 = 35. Correct again.

So I think this is the correct factorization. The key was grouping the terms in pairs where each pair shares a common factor, leading to a common binomial factor that can then be factored out. All roads lead to Rome here, so to speak. Different groupings, same result. This reinforces that the factored form is indeed (a + 5)(b + 2).
Therefore, the factored form of the expression $ab + 5b + 2a + 10$ is $\boxed{(a + 5)(b + 2)}$.
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