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anonymous

  • 4 years ago

I'm starting to forget this factoring thing, where the coefficient is not 1. can some one please go in details and explain?

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  1. precal
    • 4 years ago
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    I will show you a trick that works

  2. precal
    • 4 years ago
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    suppose \[4x^2-16x+15\]

  3. precal
    • 4 years ago
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    Take the 4 and multiply it to the 15 and get 60 rewrite as \[x^2-16x+60\] Factor (x-6)(x-10) since you took out the 4 by multiplying it, you must return it by dividing it (x-(6/4))(x-(10/4)) Whatever number you remove, you must return in (x-(3/2))(x-(5/2)) I just reduced the fractions since 3/2 is 1.5 I need to move the 2 to the front of the x since 5/2 is 2.5 I need to move the 2 to the front of the x (2x-3)(2x-5) You can check it by multiplying it out

  4. anonymous
    • 4 years ago
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    precal thank you very much.. you sir are a life sever

  5. precal
    • 4 years ago
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    I love to use this trick, it always works. Don't forget to divide the number back in...

  6. anonymous
    • 4 years ago
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    what about having big numbers please

  7. precal
    • 4 years ago
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    still works but if you have big numbers, you can always factor a number out example 4x^2+2x+2 or act like I actually used big numbers 2(2x^2+1x+1) then do the trick inside

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