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  1. #1
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    Cubics fun

    There are two parts for this question.

    1. show that, for any constant a and any natural number n, x - a is a factor of x^n - a^n.
    2. find conditions(if any) on n that are required in order that:
    i. x + a is a factor of x^n - a^n
    ii. x + a is a factor of x^n - a^n

    this is probably a big ask but I really really want to understand the theory behind this so I really need to get this properly explained!

    thanks in advance!!
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  2. #2
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    Quote Originally Posted by delicate_tears View Post
    There are two parts for this question.

    1. show that, for any constant a and any natural number n, x - a is a factor of x^n - a^n.
    Consider:

    x^{n-1}+x^{n-2}a+ \ ... \ +a^{n-2}x+a^{n-1}

    Multiply this by x to get:

    x^{n}+x^{n-1}a+ \ ... \ +a^{n-2}x^2+a^{n-1}x

    Now instead multiply it by a:

    x^{n-1}a+x^{n-2}a^2+ \ ... \ +a^{n-1}x+a^{n}

    Now if we subtract these last two expressions all but the first term of the first and the last term of the second cancell out, so:

    (x-a)(x^{n-1}+x^{n-2}a+ \ ... \ +a^{n-2}x+a^{n-1})=x^n-a^n.

    Which proves that x-a is a factor of x^n-a^n

    (Please clarrify question 2 as the two parts are the same at present).

    RonL
    Last edited by CaptainBlack; April 13th 2008 at 04:54 AM.
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  3. #3
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    Quote Originally Posted by delicate_tears View Post
    There are two parts for this question.

    1. show that, for any constant a and any natural number n, x - a is a factor of x^n - a^n.
    2. find conditions(if any) on n that are required in order that:
    i. x + a is a factor of x^n - a^n
    ii. x + a is a factor of x^n - a^n Mr F queries: Do you mean x^n + a^n?

    this is probably a big ask but I really really want to understand the theory behind this so I really need to get this properly explained!

    thanks in advance!!
    Alternatively to the good Captain:

    1. Let p(x) = x^n - a^n. Note that p(a) = 0. By the factor theorem it follows that x - a is a factor.

    2. i. x + a will only be a factor if p(-a) = 0 .....
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Alternatively to the good Captain:

    1. Let p(x) = x^n - a^n. Note that p(a) = 0. By the factor theorem it follows that x - a is a factor.

    2. i. x + a will only be a factor if p(-a) = 0 .....
    I must say that I assumed that the student had not yet done the factor theorem (since otherwise this all seems a bit too easy) - but then what do I know?

    Also the second part of the question becomes quite interesting without the factor theorem.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    I must say that I assumed that the student had not yet done the factor theorem (since otherwise this all seems a bit too easy) - but then what do I know?

    [snip]
    Well, you know what they say .... Never underestimate how difficult something easy can be.

    Quote Originally Posted by CaptainBlack View Post
    [snip]

    Also the second part of the question becomes quite interesting without the factor theorem.

    RonL
    Agreed.
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