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
(3a-5)(4a+1)
Factor: 12a^2-17a-5
2 answers
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^2-17a-5
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)
(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^2-17a-5
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)