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treydwg

  • 2 years ago

Can I factor this by grouping?

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  1. treydwg
    • 2 years ago
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    25b^2+10b+1?

  2. treydwg
    • 2 years ago
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    Can I factor this by grouping? 25b^2+10b+1?

  3. chmvijay
    • 2 years ago
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    5b+1

  4. chmvijay
    • 2 years ago
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    b=-1/5

  5. whpalmer4
    • 2 years ago
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    Prime factors of 25 are 5 and 5 so we'll have (5x + a)(5x+b). Multiply that out and you get \[(5x+a)(5x+b) = 25x^2 + 5ax + 5bx + ab\] To make that look like your equation, a and b have to = 1.

  6. rizwan_uet
    • 2 years ago
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    25b^2+10b+1 multiply 25 with 1 you will get 25. now write 10 as the sum of two numbers such that the addition of those numbers result in 10 and the product of those numbers result in 25

  7. treydwg
    • 2 years ago
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    I got b(5b+1) 1(b+1) but that does not look correct?

  8. whpalmer4
    • 2 years ago
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    No, (5b+1)(5b+1) is what you want.

  9. chmvijay
    • 2 years ago
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    (5b+1) and (5b+1) you correct it once again

  10. treydwg
    • 2 years ago
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    5b+1)^2?

  11. nitz
    • 2 years ago
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    yes...

  12. treydwg
    • 2 years ago
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    Thanks

  13. treydwg
    • 2 years ago
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    t(4t+1)=0? I got t=0, -1/4

  14. nitz
    • 2 years ago
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    ya

  15. treydwg
    • 2 years ago
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    Alright cool! Thanks

  16. nitz
    • 2 years ago
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    welcome......

  17. Hero
    • 2 years ago
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    Well, no one really did it the correct way.

  18. rizwan_uet
    • 2 years ago
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    25b^2+10b+1 25b^2 +5b+5b+1 5b(5b+1)+1(5b+1) (5b+1)(5b+1) (5b+1)^2

  19. whpalmer4
    • 2 years ago
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    No, some of us did it in a way that shows why grouping works :-)

  20. chmvijay
    • 2 years ago
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    25b25b+5b+1=0 5(5b+1)+1(5b+1)=0 (5b+1)2=0 5b+1=0 5b=-1 b=-1/5

  21. Hero
    • 2 years ago
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    Yes, but you forgot to show the process, which is also important as well @whpalmer4. The question was "How do you factor by grouping?". Showing why it works is fine. Showing the process is equally important.

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