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Jeans123 Group Title

I need help!!!!!!!!!!!!!!! Factor the expression completely, 6(x+5)^2-5(x+5)+1

  • one year ago
  • one year ago

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  1. dopek.i.d.s Group Title
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    6x(x+5)-(5x(+5))+1=6x(x+5)

    • one year ago
  2. ChmE Group Title
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    @dopek.i.d.s is incorrect

    • one year ago
  3. Jeans123 Group Title
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    I figured... level 48 and 18 at helping people hmmm..:)

    • one year ago
  4. chloe123 Group Title
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    i got to 6x^2 + 55x + 126

    • one year ago
  5. yummydum Group Title
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    umm....its completely factored but i think you mean you want it simplified...? :)

    • one year ago
  6. ChmE Group Title
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    x^2 = x*x (x+a)^2 = (x+a)(x+a) c(x+a)= c*x + c*a

    • one year ago
  7. yummydum Group Title
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    after that i think it can be factored :)

    • one year ago
  8. Jeans123 Group Title
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    Well I know the answer it is (2x+9)(3x+14)

    • one year ago
  9. yummydum Group Title
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    yes so lets see how to do that

    • one year ago
  10. Jeans123 Group Title
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    my teacher gave that to me but I do not know how to get it.

    • one year ago
  11. dopek.i.d.s Group Title
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    u guys go

    • one year ago
  12. ChmE Group Title
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    So we need to simplify it, combine like terms, then re-factor it

    • one year ago
  13. dopek.i.d.s Group Title
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    i feel stupid now

    • one year ago
  14. ChmE Group Title
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    6(x+5)^2-5(x+5)+1 6(x^2+10x+25)-5x-25+1 6x^2+55x+126

    • one year ago
  15. ChmE Group Title
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    This is where it gets difficult to factor.

    • one year ago
  16. Jeans123 Group Title
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    x=-b+-_/-b^2-4ac/2a?

    • one year ago
  17. Jeans123 Group Title
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    _/-= The square root of.

    • one year ago
  18. Jeans123 Group Title
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    Okay:)

    • one year ago
  19. yummydum Group Title
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    \[6(x+5)^2-5(x+5)+1\]lets simplify this: expand the \((x+5)\) and distribute the \(5\) to the parenthesis\[6(x^2+10x+25)-5x-25+1\]distribute the \(6\)\[6x^2+60x+150-5x-25+1\]combine like terms:\[6x^2+55x+126\]Now , we can factor it...i would use Quadratic equation since i find ti to be the "safest":\[{-55\pm\sqrt{55^2-4(6)(126)}}\over2(6)\]\[{-55\pm\sqrt{3025-3024}}\over12\]\[{-55\pm\sqrt{1}}\over12\]\[{-55\pm1}\over12\]\[{-56\over12}~~~~or~~~~{-54\over12}\]\[x=-{14\over3}~~~~or~~~~x=-{9\over2}\] so that means that its...\[(x+{14\over3})~~~and~~~(x+{9\over2})\]so to make those 'un-fractioned we'll multiply by the denominators:\[\large(3x+14)(2x+9)\] tada! :D

    • one year ago
  20. yummydum Group Title
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    @Jeans123 hope that helps! :)

    • one year ago
  21. ChmE Group Title
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    I have been proved wrong. @yummydum that was smart to use the quadratic. He found the solutions for x so he knew what the other terms had to be. Then he got rid of the fractions by multiplying by the denominator

    • one year ago
  22. yummydum Group Title
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    she* ;)

    • one year ago
  23. ChmE Group Title
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    Sry just habbit of typing, didn't mean anything by it

    • one year ago
  24. Jeans123 Group Title
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    Thank you both of you he and she;)

    • one year ago
  25. yummydum Group Title
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    no harm :) ...Youre Welcome Austin! :) feel free to ask more anytime you need help (when im there of course :))

    • one year ago
  26. Jeans123 Group Title
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    In about five minutes I will:)

    • one year ago
  27. yummydum Group Title
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    i leave pretty soon so ask now if you can :)

    • one year ago
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