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October 24, 2014

October 24, 2014

Posted by **Olivia** on Tuesday, October 2, 2012 at 9:45pm.

- Algebra II -
**Steve**, Wednesday, October 3, 2012 at 10:44amthe 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)

- Algebra II -
**Jennifer**, Wednesday, October 3, 2012 at 10:54amThis 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)

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