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Math Help - Factor theorem

  1. #1
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    Post Factor theorem

    Hi, I having trouble with this problem x^4-1 it saids to factor this problem as much as possible. Can some one please help me with this problem?
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    Quote Originally Posted by scrible View Post
    Hi, I having trouble with this problem x^4-1 it saids to factor this problem as much as possible. Can some one please help me with this problem?
    First note that it can be written as (x^2)^2 - 1^2 ....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    First note that it can be written as (x^2)^2 - 1^2 ....
    Can it be expanded into a polynomial so that I can work with it from there?
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  4. #4
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    Quote Originally Posted by scrible View Post
    Can it be expanded into a polynomial so that I can work with it from there?
    Have you been taught how to factorise the difference of two squares ....?
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  5. #5
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    Quote Originally Posted by scrible View Post
    Hi, I having trouble with this problem x^4-1 it saids to factor this problem as much as possible. Can some one please help me with this problem?
    Method 1: f(x) = x^4-1

    f(1) = (1)^4-1 = 1-1 = 0, therefore (x-1) is a factor of x^4-1.
    f(-1) = (-1)^4-1= 1-1 = 0, therefore (x+1) is a factor of x^4-1.

    So we have x^4-1 = (x-1)(x+1)(ax^2+bx+c)
    (x^2-1)(ax^2+bx+c) = x^4-1
    x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1
    ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1
    (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1

    \begin{cases} a = 1 \\<br />
 b = 0 \\<br />
 c-a = 0 & c-1  = 0 \implies c = 1 \\<br />
-b = 0 \\<br />
-c = -1 &  c = 1. \end{cases}

    Therefore ax^2+bx+c = x^2+1

    Hence \boxed{(x^2-1)(x^2+1) = x^4-1}

    Method 2: Any products yielding answers of the form a^n-b^n are given by:

    (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-1}+b^{n-1}) = a^n-b^n

    So for x^4-1^2, we have:

    (x-1)((x^{4-1})+(x^{4-2})({1})+(x^{4-3})({1})^2+(x^{4-4})({1})^3)= x^4-1^2

    (x-1)((x^{3})+(x^{2})({1})+(x^{1})({1})^2+(x^{0})({1}  )^3)= x^4-1^2

    (x-1)(x^3+x^2+x+1) = x^4-1

    x^3+x^2+x+1 yields to zero when x = -1, so, by the factor theorem:

    (x+1)(x-1)(ax^2+bx+c) = x^4-1

    (x^2-1)(ax^2+bx+c) = x^4-1
    x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1
    ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1
    (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1

    \begin{cases} a = 1 \\<br />
 b = 0 \\<br />
 c-a = 0 & c-1  = 0 \implies c = 1 \\<br />
-b = 0 \\<br />
-c = -1 &  c = 1. \end{cases}

    Therefore ax^2+bx+c = x^2+1

    Hence \boxed{(x^2-1)(x^2+1) = x^4-1}
    Last edited by I4talent; November 1st 2009 at 11:47 PM.
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  6. #6
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    I appreciate the time and effort you have put into this reply. However, when someone gives hints on how to solve a problem, please do not then give a full solution; it is forum etiquette to wait until the original poster has responded before replying. (And I suspect that the path I'm trying to steer the OP down is a more accessible one).

    Quote Originally Posted by I4talent View Post
    Method 1: f(x) = x^4-1

    f(1) = (1)^4-1 = 1-1 = 0, therefore (x-1) is a factor of x^4-1.

    f(-1) = (-1)^4-1= 1-1 = 0, therefore (x+1) is a factor of x^4-1.

    So we have x^4-1 = (x-1)(x+1)(ax^2+bx+c)

    (x^2-1)(ax^2+bx+c) = x^4-1

    x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1

    ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1

    (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1

    \begin{cases} a = 1 \\<br /> <br />
b = 0 \\<br /> <br />
c-a = 0 & c-1 = 0 \implies c = 1 \\<br /> <br />
-b = 0 \\<br /> <br />
-c = -1 & c = 1. \end{cases}

    Therefore ax^2+bx+c = x^2+1

    Hence \boxed{(x^2-1)(x^2+1) = x^4-1}



    Method 2: Any products yielding answers of the form a^n-b^n are given by:

    (a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-1}+b^{n-1}) = a^n-b^n

    So for x^4-1^2, we have:

    (x-1)((x^{4-1})+(x^{4-2})({1})+(x^{4-3})({1})^2+(x^{4-4})({1})^3)= x^4-1^2

    (x-1)((x^{3})+(x^{2})({1})+(x^{1})({1})^2+(x^{0})({1}  )^3)= x^4-1^2

    (x-1)(x^3+x^2+x+1) = x^4-1

    x^3+x^2+x+1 yields to zero when x = -1, so, by the factor theorem:

    (x+1)(x-1)(ax^2+bx+c) = x^4-1

    (x^2-1)(ax^2+bx+c) = x^4-1

    x^2(ax^2+bx+c)-1(ax^2+bx+c) = x^4-1

    ax^4+bx^3+cx^2-ax^3-bx^2-cx = x^4-1

    (a)x^4+(b)x^3+(c-a)x^2-(b)x-c = x^4-1

    \begin{cases} a = 1 \\<br /> <br />
b = 0 \\<br /> <br />
c-a = 0 & c-1 = 0 \implies c = 1 \\<br /> <br />
-b = 0 \\<br /> <br />
-c = -1 & c = 1. \end{cases}

    Therefore ax^2+bx+c = x^2+1

    Hence \boxed{(x^2-1)(x^2+1) = x^4-1}
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    I appreciate the time and effort you have put into this reply. However, when someone gives hints on how to solve a problem, please do not then give a full solution; it is forum etiquette to wait until the original poster has responded before replying. (And I suspect that the path I'm trying to steer the OP down is a more accessible one).
    I'm sorry. I was not aware of that. But does this rule - or convention - apply only when someone has given a hint, or does everybody have to give hints and not the whole solution unless it's the last resort?
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  8. #8
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    Quote Originally Posted by I4talent View Post
    I'm sorry. I was not aware of that. But does this rule - or convention - apply only when someone has given a hint, or does everybody have to give hints and not the whole solution unless it's the last resort?
    Giving hints and nudges in the right direction is usually preferable to giving a detailed and complete solution.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Giving hints and nudges in the right direction is usually preferable to giving a detailed and complete solution.
    Thank you I got now.
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