Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

factor 20x^2-23x-7....?

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  • iceman
    Lv 7
    8 years ago
    Favorite Answer

    20x^2-23x-7

    = (4x+1) (5x-7)

    Edit: adding explanation: find two numbers that multiply to -140 , add to - 23:

    20x^2 + 5x - 28x - 7

    = 5x(4x + 1) - 7(4x + 1)

    = (4x + 1)(5x - 7)

  • 8 years ago

    You know you need the "(blah)(blah)". So to find the two numbers to factor it by, multiply A by C, the standard equation is Ax2+Bx+C. which is -140, then see what two numbers add up to -23 and also multiply to -140. They are:

    -1 140 or 1 -140

    -2 70 2 -70

    -5 28 5 -28

    -7 20 7 -20

    -10 14 10 -14

    Then see which one of those add up to -23, which is 5 and -28.

    It is then 20x^2 +5x -28x -7

    Then (20x^2+5x)(-28x-7)

    take out common factors

    5x(4x+1)-7(4x+1)

    The (4x+1) become one, that's hoe I learned it and the 5x-7 go into another one, (5x-7).

    Then the final answer is:

    (5x-7)(4x+1)

    Source(s): Algebra II Honors, I remember this lesson from last year.
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