Posted by Olivia on Tuesday, October 2, 2012 at 9:45pm.
- Algebra II - Steve, Wednesday, October 3, 2012 at 10:44am
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
- Algebra II - Jennifer, Wednesday, October 3, 2012 at 10:54am
This will be factored into an expression of the form
(A*a + B)(C*a + D)
A*C*a^2 + A*D*a + B*C*a + B*D =
A*C*a^2 + (A*D + B*C)*a + B*D =
A*C = 12
A*D + B*C = -17
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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