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treydwg

  • one year ago

Can I factor this by grouping?

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

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

  3. chmvijay
    • one year ago
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    5b+1

  4. chmvijay
    • one year ago
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    b=-1/5

  5. whpalmer4
    • one year 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
    • one year 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
    • one year ago
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    I got b(5b+1) 1(b+1) but that does not look correct?

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

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

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

  11. nitz
    • one year ago
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    yes...

  12. treydwg
    • one year ago
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    Thanks

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

  14. nitz
    • one year ago
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    ya

  15. treydwg
    • one year ago
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    Alright cool! Thanks

  16. nitz
    • one year ago
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    welcome......

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

  18. rizwan_uet
    • one year 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
    • one year ago
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    No, some of us did it in a way that shows why grouping works :-)

  20. chmvijay
    • one year 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
    • one year 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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