Post a New Question

Algebra II

posted by .

Factor: 12a^2-17a-5

  • Algebra II -

    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)

  • Algebra II -

    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)

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question