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anonymous

  • 4 years ago

How do I solve: 4y^2 - 11y - 3 =0 by factoring?

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  1. anonymous
    • 4 years ago
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    Well, do you know how to factor by decomposition?

  2. anonymous
    • 4 years ago
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    by decomposition?

  3. anonymous
    • 4 years ago
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    Yeah, we compose the middle term into smaller terms

  4. anonymous
    • 4 years ago
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    decompose*

  5. anonymous
    • 4 years ago
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    yes I do

  6. anonymous
    • 4 years ago
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    but if I do that: (2y-...)(2y-...)

  7. anonymous
    • 4 years ago
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    So what two numbers sum up to -11, and are multiplied to get -12?

  8. anonymous
    • 4 years ago
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    to get -12?

  9. anonymous
    • 4 years ago
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    Yeah?

  10. anonymous
    • 4 years ago
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    ohhh, 4 x -3

  11. anonymous
    • 4 years ago
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    i get it now

  12. anonymous
    • 4 years ago
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    Its 12, and minus one, it has to also add up to -11

  13. anonymous
    • 4 years ago
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    yup! okay, thank you!

  14. anonymous
    • 4 years ago
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    Your welcome :), ill be here for a bit if you need anymore help

  15. anonymous
    • 4 years ago
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    You can factor by grouping... \[4y^2-11y-3=0\] \[4y^2-12y+y-3=0\] Remove a 3 from the first 2 terms. \[4y(y-3)+(y-3)=0\] Factor out a (y-3) \[(y-3)(4y+1)=0\] Easy to find the zeroes from here.

  16. anonymous
    • 4 years ago
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    Remove a 4y from the first 2 terms***

  17. anonymous
    • 4 years ago
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    Why do I have to remove a 4y from the first 2 terms? And why is it now (y-3)(4y + 1)?

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