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Math Help - Confusing Question

  1. #1
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    Confusing Question

    Can somebody please explain to me what is meant by the following question:

    If (8x-5)^5=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F, find A+B+C+D+E+F.

    Thanks.
    Last edited by drguildo; January 30th 2009 at 03:55 PM.
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  2. #2
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    (8x-5)^5=(8x-5)(8x-5)(8x-5)(8x-5)(8x-5)

    Multiply the parantheseses and you should get the constants (A,B,C....) infront of each x^n value where n=0,1,2,3,4,5.

    Hope I helped you somewhat
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  3. #3
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    You can use Binomial Theorem/Pascal's Triangle to expand. Look them up if you're not familiar with it.

    To start you off:
    (8x+(-5))^5 = (8x)^5+5(8x)^4(-5)+10(8x)^3(-5)^2+10(8x)^2(-5)^3+5(8x)(-5)^4 +(-5)^5
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  4. #4
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    I'm more confused than when I posted the question. It's like I'm so confused that I don't even know what I'm confused about.

    The question is from a book on basic algebra. The point the question is asked is before multiplying out parenthesis has been covered.

    The answer states:

    Letting x = 1 one gathers that 243 = 3^5 = (8(1) - 5)^5 = A+B+C+D+E+F.

    This doesn't make any sense to me. Why 1? Why not 20: 89466096875 = 155^5 = (8(20) - 5)^5 = A+B+C+D+E+F

    Should I learn from a different book? This is not the first time there's been a question like this.
    Last edited by drguildo; January 30th 2009 at 05:00 PM. Reason: I am stupid.
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  5. #5
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    Oh wow I can't believe I went about it methodically without thinking a bit more.

    (8x-5)^5=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F

    This equality is supposed to hold for all x. If we input x = 1, we managed to make the right side into the expression in question. The left side would result in:
    (8(1)-5)^5 = (3)^5 = 243

    Nah, your book is fine so far. You did not need to multiply nor did you need Pascal's triangle *slaps forehead*
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