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anonymous

  • one year ago

Someone please explain to me the different methods of factoring a trinomial when the leading coefficient isn't one. Ex: 6x^2+31x-30 = 0

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  1. anonymous
    • one year ago
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    You can use the quadratic formula.

  2. anonymous
    • one year ago
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    \[\huge \frac{ -b \pm \sqrt{b^2-4ac} }{2a }\]

  3. anonymous
    • one year ago
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    if i were to use the quadratic formula wouldn't I get a fairly large number under the radical though? isn't there some way to avoid that?

  4. anonymous
    • one year ago
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    where \(\huge ax^2+bx+c\). No, you wouldn't as long as you know how to solve it.

  5. anonymous
    • one year ago
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    Another approach:

  6. anonymous
    • one year ago
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    earlier I had a question where the number under the radical ended up being something like 1137. In that scenario what route would I take?

  7. anonymous
    • one year ago
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    Another approach: factor it.

  8. anonymous
    • one year ago
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    Though it may seem frustrating because it take so much time since it's a trial-and-error-method.

  9. anonymous
    • one year ago
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    I'm just not a fan of math.

  10. anonymous
    • one year ago
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    I'll use the quadratic formula to show to you that it's not hard. :)

  11. anonymous
    • one year ago
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    So we have \(6x^2+31x-30=0\) where \(a=6\) \(b=31\) and \(c=-30\).

  12. anonymous
    • one year ago
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    THE QUADRATIC FORMULA: \(\huge \frac{ -b \pm \sqrt{b^2-4ac} }{2a }\) PLUG IN THE VALUES. \(\huge \frac{ -31 \pm \sqrt{31^2-4(6)(-30)} }{2(6) }\) NOW DO THE ALGEBRA. \(\huge x=\)\(\huge \frac{ -31 \pm \sqrt{961+720} }{12 }\) \(\huge x=\)\(\huge \frac{ -31 \pm \sqrt{1681} }{12 }\) \(\huge x=\)\(\huge \frac{ -31 \pm {41} }{12 }\) WE KNOW THAT THIS EQUATION HAS TWO SOLUTIONS. 1.) \(\huge x=\)\(\huge \frac{ -31 + {41} }{12 }\) 2.) \(\huge x=\)\(\huge \frac{ -31 - {41} }{12 }\)

  13. anonymous
    • one year ago
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    Now do the math. :)

  14. anonymous
    • one year ago
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    Are you now okay with this or do you still need more help?

  15. anonymous
    • one year ago
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    That works for me, thank you very much.

  16. anonymous
    • one year ago
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    Anytime! Tell me what you got and I'll say if it's right or wrong. :)

  17. anonymous
    • one year ago
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    -6, and 5/6. thank you very much.

  18. anonymous
    • one year ago
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    You're right! :) You're welcome!

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