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anonymous

  • 4 years ago

how can i factor this out? 28b^2-80b-128

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  1. mathteacher1729
    • 4 years ago
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    First ask yourself if there is a constant you can factor out of all the coefficients. :)

  2. anonymous
    • 4 years ago
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    4

  3. anonymous
    • 4 years ago
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    right?

  4. mathteacher1729
    • 4 years ago
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    Yup. So now write 4 times (whatever is left over after you factor out 4)

  5. anonymous
    • 4 years ago
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    4(7b^2-20b-32) but then what do i do?

  6. mathteacher1729
    • 4 years ago
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    Now we want to factor out 7b^2-20b-32 into the product of two binomaisl: (b plus or minus stuff) times ( b plus or minus stuff). BUT LOOK! The coefficient of b^2 is 7, so that means we automatically know that it's going to look like... (7b plus or minus stuff) times (b plus or minus stuff).

  7. anonymous
    • 4 years ago
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    well that is where i'm confused

  8. anonymous
    • 4 years ago
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    how do i know what number to choose?

  9. mathteacher1729
    • 4 years ago
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    Can you factor something like: x^2 + 10x +21 ? It's similar to that, but just a little harder.

  10. anonymous
    • 4 years ago
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    but if i factor out -32 there is nothing that would match -20

  11. anonymous
    • 4 years ago
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    is there anything i can do?

  12. mathteacher1729
    • 4 years ago
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    The answer to this problem is the same as factoring something like x^2 + 10x +21 Describe how you would factor this and then the solution to your problem will make much more sense. :)

  13. anonymous
    • 4 years ago
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    is it (x+7)(x+3) ?

  14. mathteacher1729
    • 4 years ago
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    Yes! :) How did you obtain that answer? How did you know to go for +7 and +3? Also, can you factor: x^2 -4x -21

  15. anonymous
    • 4 years ago
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    well because the numbers are positive But i'm confused w/ my original problem

  16. mathteacher1729
    • 4 years ago
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    7b^2-20b-32 = (7b ) * ( b ) The two numbers you fill in will have to multiply to negative 32. That's the start of it.

  17. anonymous
    • 4 years ago
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    right but there is 1*32,2*16,4*8

  18. mathteacher1729
    • 4 years ago
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    You have to try those out. :) ALSO note how one of those numbers is gonna be multplied by 7 when you FOIL. So it's trial and error. After a bit of practice you develop a number sense about these things. So for instance 1 * 32 won't work... (7b +32)(b -1) FOIL that out and you get 7b^2 32b -7b = -25b -31 So try the next one...

  19. anonymous
    • 4 years ago
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    okay is it 4 and 8

  20. mathteacher1729
    • 4 years ago
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    In what order? And with what signs (which is + and which is - ) ?

  21. anonymous
    • 4 years ago
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    (7b-4)(b+8)

  22. mathteacher1729
    • 4 years ago
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    Let's factor that out now: (7b-4)(b+8) 7b^2 (good) -4b +56b = 52b (nope) 8 * -4 = -32 (good)

  23. mathteacher1729
    • 4 years ago
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    You are VERY close though!

  24. anonymous
    • 4 years ago
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    i don't know where to go from there

  25. anonymous
    • 4 years ago
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    i'm stuck

  26. anonymous
    • 4 years ago
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    is it (7b-1)(b+2) ?

  27. mathteacher1729
    • 4 years ago
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    No, because -1 * 2 = -2. You were right, it's 4 and 8. But the order and the sign need to be tweaked.

  28. anonymous
    • 4 years ago
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    what do u mean?

  29. mathteacher1729
    • 4 years ago
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    (7b )(b ) Fill in the blanks with 4 and 8. One of the combinations will work. :)

  30. mathteacher1729
    • 4 years ago
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    Do you know how to FOIL? (or expand) a binomial? That is the other half of solving problems like these. Going from (7b +4)(b +8) to 7b^2 +12b +32 is critical to the trial-and-error portion.

  31. mathteacher1729
    • 4 years ago
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    Paul's Online Math Notes has a very nice explanation of this with TONS of examples: http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx I have to sign off, but hopefully this will help a little. :)

  32. anonymous
    • 4 years ago
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    okay thank you!

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