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Jeans123

  • 3 years ago

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

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  1. dopek.i.d.s
    • 3 years ago
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    6x(x+5)-(5x(+5))+1=6x(x+5)

  2. ChmE
    • 3 years ago
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    @dopek.i.d.s is incorrect

  3. Jeans123
    • 3 years ago
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    I figured... level 48 and 18 at helping people hmmm..:)

  4. chloe123
    • 3 years ago
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    i got to 6x^2 + 55x + 126

  5. yummydum
    • 3 years ago
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    umm....its completely factored but i think you mean you want it simplified...? :)

  6. ChmE
    • 3 years ago
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    x^2 = x*x (x+a)^2 = (x+a)(x+a) c(x+a)= c*x + c*a

  7. yummydum
    • 3 years ago
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    after that i think it can be factored :)

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

  9. yummydum
    • 3 years ago
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    yes so lets see how to do that

  10. Jeans123
    • 3 years ago
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    my teacher gave that to me but I do not know how to get it.

  11. dopek.i.d.s
    • 3 years ago
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    u guys go

  12. ChmE
    • 3 years ago
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    So we need to simplify it, combine like terms, then re-factor it

  13. dopek.i.d.s
    • 3 years ago
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    i feel stupid now

  14. ChmE
    • 3 years ago
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    6(x+5)^2-5(x+5)+1 6(x^2+10x+25)-5x-25+1 6x^2+55x+126

  15. ChmE
    • 3 years ago
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    This is where it gets difficult to factor.

  16. Jeans123
    • 3 years ago
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    x=-b+-_/-b^2-4ac/2a?

  17. Jeans123
    • 3 years ago
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    _/-= The square root of.

  18. Jeans123
    • 3 years ago
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    Okay:)

  19. yummydum
    • 3 years ago
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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

  20. yummydum
    • 3 years ago
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    @Jeans123 hope that helps! :)

  21. ChmE
    • 3 years ago
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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

  22. yummydum
    • 3 years ago
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    she* ;)

  23. ChmE
    • 3 years ago
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    Sry just habbit of typing, didn't mean anything by it

  24. Jeans123
    • 3 years ago
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    Thank you both of you he and she;)

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

  26. Jeans123
    • 3 years ago
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    In about five minutes I will:)

  27. yummydum
    • 3 years ago
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    i leave pretty soon so ask now if you can :)

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