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The distributive property says
A·(B + C) = A·B + A·C
But it also says the reverse
A·B + A·C = A·(B + C)
Write 13 as 13·1 using the identity property of multiplication
and write 13x as 13·x.
13·1 + 13·x = 13·(1 + x)
That is
A·B + A·C = A·(B + C)
↓ ↓ ↓ ↓ ↓ ↓ ↓
13·1 + 13·x = 13·(1 + x)
with 13 substituted for A, 1 substituted for B and x substituted for x.
Edwin