Algebra II
posted by Olivia .
Factor: 12a^217a5

the factorization will be one of
(12a )(a+ )
(12a+ )(a )
(6a )(2a+ )
(6a+ )(2a )
(3a )(4a+ )
(3a+ )(4a )
where the missing numbers are 1 and 5
a little trial produces
(3a5)(4a+1) 
This will be factored into an expression of the form
(A*a + B)(C*a + D)
Multiplying out:
A*C*a^2 + A*D*a + B*C*a + B*D =
A*C*a^2 + (A*D + B*C)*a + B*D =
12a^217a5
which means
A*C = 12
A*D + B*C = 17
B*D =5
Then start trying whole numbers. . .B and D have to be +/ 1 or +/5
A and C have to be positive common factors of 12: so 1 and 12 or 3 and 4 or 6 and 2. . .
4*5 + 3*1 = 17. . .
A = 4, B = 1, C = 3, D = 5
The factored expression is
(4a + 1)*(3a  5)
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