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Math Help - Factoring and impossible equation!!!

  1. #1
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    Factoring and impossible equation!!!

    (20b(squared)-11b-3)
    Sorry I don't know how to type the squared symbol...but thanks in advance!
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by curlysue View Post
    (20b(squared)-11b-3)
    Sorry I don't know how to type the squared symbol...but thanks in advance!
    20b^2 -11b -3= (4b -3)(5b+1)
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by curlysue View Post
    (20b(squared)-11b-3)
    Sorry I don't know how to type the squared symbol...but thanks in advance!
    20b^2 - 11b - 3

    Multiply the coefficient of the leading term by the constant term:
    20 * -3 = -60

    List all the pairs of factors of -60:
    1, -60
    2, -30
    3, -20
    4, -15
    5, -12
    6, -10
    10, -6
    12, -5
    15, -4
    20, -3
    30, -2
    60, -1

    Now we want to find the pair of factors in the list above that add up to be equal to the coefficient of the linear term, in this case -11. (If no such pair of factors is in the list then you cannot factor the quadratic.)
    I've got 4 + -15 = -11.

    So split the linear term of the quadratic into -11x = 4x - 15x.
    20b^2 - 11b - 3 = 20b^2 + 4b - 15b - 3

    Group the terms as follows:
    = (20b^2 + 4b) + (-15b - 3)

    Now factor from each group:
    = 4b(5b + 1) + (-3)(5b + 1)

    and note that both terms have a common factor, 5b + 1, so we can factor that from each term as well:
    = (4b - 3)(5b + 1)

    There you go!

    -Dan
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by topsquark View Post
    20b^2 - 11b - 3

    Multiply the coefficient of the leading term by the constant term:
    20 * -3 = -60

    List all the pairs of factors of -60:
    1, -60
    2, -30
    3, -20
    4, -15
    5, -12
    6, -10
    10, -6
    12, -5
    15, -4
    20, -3
    30, -2
    60, -1

    Now we want to find the pair of factors in the list above that add up to be equal to the coefficient of the linear term, in this case -11. (If no such pair of factors is in the list then you cannot factor the quadratic.)
    I've got 4 + -15 = -11.

    So split the linear term of the quadratic into -11x = 4x - 15x.
    20b^2 - 11b - 3 = 20b^2 + 4b - 15b - 3

    Group the terms as follows:
    = (20b^2 + 4b) + (-15b - 3)

    Now factor from each group:
    = 4b(5b + 1) + (-3)(5b + 1)

    and note that both terms have a common factor, 5b + 1, so we can factor that from each term as well:
    = (4b - 3)(5b + 1)

    There you go!

    -Dan
    That's pretty cool, I actually didn't know about the ordered pair method.
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  5. #5
    MHF Contributor red_dog's Avatar
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    20b^2-11b-3=20b^2+4b-15b-3=4b(5b+1)-3(5b+1)=(5b+1)(4b-3)
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  6. #6
    Super Member wingless's Avatar
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    If you can't see how to factorize, you can use this:

    ax^2 + bx + c = (x - x_1)(x-x_2)

    20x^2 - 11x - 3

    x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}=\frac{11+\sqrt{361}}{40} = \frac{3}{4}

    x_2 = \frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{11-\sqrt{361}}{40} = -\frac{1}{5}

    ax^2 + bx + c = \left(x - \frac{3}{4}\right)\left(x+\frac{1}{5}\right)

    Multiplying by 4.5 gives 4.\left(x - \frac{3}{4}\right).5.\left(x+\frac{1}{5}\right)

    (4x-3)(5x+1)
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  7. #7
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by wingless View Post
    If you can't see how to factorize, you can use this:

    ax^2 + bx + c = (x - x_1)(x-x_2)
    ax^2+bx+c=a(x-x_1)(x-x_2)
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by angel.white View Post
    That's pretty cool, I actually didn't know about the ordered pair method.
    Honestly I didn't learn it myself until I taught a Pre-Calc College class. It's called the "a-c" method. (The "a-c" comes from the coefficients in ax^2 + bx + c.)

    -Dan
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