To factor the expression \(3mn + 2 + 12n + 8\) by grouping, let's first rearrange and group the terms:
- Group terms: \((3mn + 12n) + (2 + 8)\)
- Factor out the common factors in each group:
- From \(3mn + 12n\), we can factor out \(3n\): \[ 3n(m + 4) \]
- From \(2 + 8\), we factor out \(2\): \[ 2(1 + 4) \text{ which simplifies to } 10 \]
Now we rewrite the expression: \[ 3n(m + 4) + 10 \] This doesn't seem right in terms of keeping the terms that remain properly.
So I'll actually return to this step:
The original expression was: \[ 3mn + 12n + 2 + 8 \]
Let's regroup:
- \(3mn + 12n + 2 + 8 = 3n(m + 4) + 10\)
- Rearranging yields \(3n(m + 4) + 10\)
Now we can see that when we completely factor: \[ (3n + 10)(m + 4) \]
That seems incorrect; let's refine this too:
Noticing again, we purely take out one factor, we were off in grouping.
Correcting it you take proper terms:
Let's check the original group step again:
Rewrite it entirely,
\[ (3mn + 12n) + (2 + 8) \]
Now, finalize again with another step of showing it as pure factored alongside direct: \[ 3n(m + 4) + 10 \]
This aligns up with expecting a result, which was found to lead.
Looking back at your need \( (m + 4)(?+2)\):
This leads to a direct peek at expansion.
Observing closely:
The second factor in this \(? + 2\) would take the full \(n + constant\) thus matching:
So to conclude, our grouped factors showcase proper adjustment towards:
Finalized as \((m + 4)(n + 2)\) leading:
Missing term is: \[ \boxed{n} \]
So hence rewritten as compactly would derive into boundaries expected.
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