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Math Help - Factorise the Following

  1. #1
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    Factorise the Following

    Please can someone help in factorising the following quadratic using simple factors....

    20x^2+7x-6


    Thanks in advance.

    JP
    Last edited by jp.101; December 23rd 2008 at 09:37 AM.
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  2. #2
    MHF Contributor

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    Factoring 20x^2+ 7x- 6 means finding (ax+ b)(cx+ d) so that the product, acx^2+ (ad+ bc)x+ bd is equal to that. That means ac= 20, bd= -6, and ad+ bc= 7. It would be impossible to find a, b, c, and d without restricting them to integers (and might be impossible even so).

    We are looking for integers a and b so that ab= 20. Well 20= 1*20, 2*10, 4*5, (-1)*(-20), (-2)*(-10), or (-4)*(-5). We are also looking for integers c and so that bd= -6. -6= (-1)*6, (-2)*3, 1*(-6), or 2*(-3). How do we decide which? TRY them! Which of those give ad+ bc= 7?
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  3. #3
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    You can use the quadratic formula to factor. For ax^2+bx+c=0, x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}.

    Factoring 20x^{2}+7x-6=0.

    Using the quadratic formula:

    x={ \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}}= { \dfrac{-(7) \pm \sqrt{(7)^{2}-4(20)(-6)}}{2(20)}} = {\dfrac{-7 \pm 23}{40}}.

    Thus x = \dfrac{2}{5} or x = -\dfrac{3}{4}. Now take each of these solutions:

    x = \dfrac{2}{5}, multiplying both sides by 5 then subtracting 2 from both sides yields:

    (1) 5x-2 = 0.

    x = -\dfrac{3}{4}, multiplying both sides by 4 and adding 3 to both sides yields:

    (2) 4x+3=0.

    Multiplying equations (1) and (2) together we get the factorized form of 20x^{2}+7x-6=0:

    (5x-2)(4x+3)=0.
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