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Math Help - Help me factor a polynomial from graph thepry

  1. #1
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    Help me factor a polynomial from graph thepry

    Hello, I am having a hard time understanding why k(k - 1)^n - (k - 1)^n - (-1)^n(k - 1) = (k - 1)^(n+1) + (-1)^(n+1) (k - 1).
    What i specifically don't get is how to get from ^n to ^n+1. I would appreciate it if someone could tell me how this is done.
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    Re: Help me factor a polynomial from graph thepry

    Hey takaj.

    This is based on distributive law where k(k-1)^n - (k-1)^n = (k-1)^1*(k-1)^n = (k-1)^(n+1).

    This is just based on ab + cb = (a+c)*b.
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by chiro View Post
    Hey takaj.

    This is based on distributive law where k(k-1)^n - (k-1)^n = (k-1)^1*(k-1)^n = (k-1)^(n+1).

    This is just based on ab + cb = (a+c)*b.
    Hi chiro , I don't seem to understand your explanation. Could you just go a little deeper or maybe solve it line by line. That would be great!
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    Re: Help me factor a polynomial from graph thepry

    Let b = (k-1)^n, a = k and c = -1.

    We know from distributive law that (a+c)b = ab + bc if we expand out but we can do the reverse: if we have ab + bc we can write this as (a+c)b.

    Now ab = k*(k-1)^n and bc = -(k-1)^n so (a+c)b = (k + -1)*(k-1)^n = (k-1)*(k-1)^n but from powers of terms, we have z^a * z^b = z^(a+b) (exponentials) so in this case z = (k-1), a = 1 and b = n we get a+b = n+1 with our z so the final result is (k-1)^(n+1).
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by chiro View Post
    Let b = (k-1)^n, a = k and c = -1.

    We know from distributive law that (a+c)b = ab + bc if we expand out but we can do the reverse: if we have ab + bc we can write this as (a+c)b.

    Now ab = k*(k-1)^n and bc = -(k-1)^n so (a+c)b = (k + -1)*(k-1)^n = (k-1)*(k-1)^n but from powers of terms, we have z^a * z^b = z^(a+b) (exponentials) so in this case z = (k-1), a = 1 and b = n we get a+b = n+1 with our z so the final result is (k-1)^(n+1).
    ok, so how do you get from (-1)^n to (-1)^n+1?
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  6. #6
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    Re: Help me factor a polynomial from graph thepry

    Ok so your term is - (-1)^(n+1). Now this is the same as (-1)^n * (-1). But this changes your - (since this is there before the term) to a plus if you expand it out.

    So your - (-1)^n * (-1) changes the - to a + giving - - (-1)^n = + (-1)^n.
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by takaj View Post
    k(k - 1)^n - (k - 1)^n - (-1)^n(k - 1)
    I found it easier this way:

    Let a = k - 1, b = n - 1 ; then expression becomes:
    k * a^n - a^n - a * (-1)^n

    = a[k * a^b - a^b - (-1)^n]

    = a[a^b * (k - 1) - (-1)^n]

    = a[a^b * a - (-1)^n]

    = a[a^(b + 1) - (-1)^n] ; b = n - 1 :

    = a[a^n - (-1)^n]

    substitute back a = k - 1
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    Re: Help me factor a polynomial from graph thepry

    I am really not following. Could one of you recommend a website where this 'rule' is taught?
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    Re: Help me factor a polynomial from graph thepry

    What are you not following? Stuff like a * a^(n-1) = a^n ?
    If so, you need classroom help...
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by Wilmer View Post
    What are you not following? Stuff like a * a^(n-1) = a^n ?
    If so, you need classroom help...
    No, no, i understand your above example.

    But, tell me if this is correct...

    (-1)^n *(k-1) = (-1)^(n+1) * (k-1)
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  11. #11
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    Re: Help me factor a polynomial from graph thepry

    This is not correct: the only reason why you can do the above is because it changes from a - to a + before the term.
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  12. #12
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by chiro View Post
    This is not correct: the only reason why you can do the above is because it changes from a - to a + before the term.
    When the variable exponents are replaced with real numbers, I have no trouble manipulating the polynomial. But when there is a variable there, I get really confused. I don't think i need 'classroom help.'

    I understand that x^n * 1^n = x^n+n, but i just don't understand how this logic is being applied to the polynomial in the original post.

    actually, on rereading your original post, chiro, i think i get what you are saying. So you are manipulating the polynomial using the distribute law. I think i need to read up on this law. I know about it from logic, but i'm a bat hazy at applying it to numbers
    Last edited by takaj; October 1st 2012 at 05:58 AM.
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    Re: Help me factor a polynomial from graph thepry

    It's because it involves -1 to some power and it's a little trick.

    If you have - a*(-1)^n, then this is the same as + a*(-1)^(n+1): it's because of it being a -1 and not something else.
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by chiro View Post
    It's because it involves -1 to some power and it's a little trick.

    If you have - a*(-1)^n, then this is the same as + a*(-1)^(n+1): it's because of it being a -1 and not something else.
    So i take it that a =(K-1)

    That means that - (-1)^n * (k - 1) = +(-1)^n+1 * (k-1)

    Am i correct?
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    Re: Help me factor a polynomial from graph thepry

    Quote Originally Posted by takaj View Post
    I understand that x^n * 1^n = x^n+n,...
    1^(anything) = 1
    So x^n * 1^n = x^n

    RULE: a^n * b^n = (ab)^n

    ALSO: (-1)^2n = 1 ; (-1)^(2n+1) = -1 ; where n is any integer
    Last edited by Wilmer; October 1st 2012 at 06:20 AM.
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