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  1. #1
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    polynomials

    Find a and b (from reals) so that

    $\displaystyle p(x) = ax^{2008} + bx^{2007} - 1$
    will be dividable by polynomial $\displaystyle (x - 1)^2$.

    no idea how to begin.. can someone help?
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  2. #2
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    Quote Originally Posted by metlx View Post
    Find a and b (from reals) so that

    $\displaystyle p(x) = ax^{2008} + bx^{2007} - 1$
    will be dividable by polynomial $\displaystyle (x - 1)^2$.

    no idea how to begin.. can someone help?
    i think $\displaystyle P$ will be dividable by $\displaystyle (x-1)^2$ if $\displaystyle P(1) =0.$
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Raoh View Post
    i think $\displaystyle P$ will be dividable by $\displaystyle (x-1)^2$ if $\displaystyle P(1) =0.$
    Yes, but I believe that this is only half of the story. 1 is a double root.
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  4. #4
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    so how do i solve this?

    I did this:

    If $\displaystyle p(1) = 0 \Rightarrow a + b = 1$
    $\displaystyle a = 1 - b$

    $\displaystyle x_{1,2} = \frac{-b \pm \sqrt{b^2 -(- 4(1 - b))}}{2(1 - b)} = \frac{-b \pm \sqrt{b^2 - 4b + 4}}{2 - 2b}$

    $\displaystyle = \frac {-b \pm \sqrt{(b - 2)^2}}{2 - 2b} = \frac{-b \pm (b - 2)}{2 - 2b}$

    $\displaystyle x_1 = \frac{-b + b - 2}{2 - 2b} = - \frac {1}{1 - b} = - \frac {1}{a}$

    $\displaystyle x_2 = \frac{ -b - b + 2}{2 - 2b} = 1$

    does this help me in anyway?

    i can put those values instead of x-es..
    $\displaystyle a \cdot (-\frac{1}{a})^{2008} + b \cdot (- \frac{1}{1 - b})^{2007} - 1 = 0$

    $\displaystyle a \cdot \frac{1}{a} - \frac{b}{1 - b} - 1 = 0 \Rightarrow -\frac{b}{1 - b} = 0 \Rightarrow b = 0$

    a = 1 - b = 1.

    is that how you do it?
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  5. #5
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    Quote Originally Posted by VonNemo19 View Post
    Yes, but I believe that this is only half of the story. 1 is a double root.
    try to use the synthetic division (Synthetic division - Wikipedia, the free encyclopedia) to divide $\displaystyle p(x)$ by $\displaystyle x^2 - 2x + 1$

    hope it'll work
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  6. #6
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    Quote Originally Posted by dedust View Post
    try to use the synthetic division (Synthetic division - Wikipedia, the free encyclopedia) to divide $\displaystyle p(x)$ by $\displaystyle x^2 - 2x + 1$

    hope it'll work
    is that the same as Horner's algoritm?
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  7. #7
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    Quote Originally Posted by VonNemo19 View Post
    Yes, but I believe that this is only half of the story. 1 is a double root.
    exactly.
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    Quote Originally Posted by metlx View Post
    is that the same as Horner's algoritm?
    yes, with a slight modification
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    Quote Originally Posted by dedust View Post
    try to use the synthetic division (Synthetic division - Wikipedia, the free encyclopedia) to divide $\displaystyle p(x)$ by $\displaystyle x^2 - 2x + 1$

    hope it'll work
    the degree of P is a LOT bigger than $\displaystyle x^2 - 2x + 1$,are you really sure about doing the division.
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  10. #10
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    Quote Originally Posted by Raoh View Post
    exactly.
    so did i do it right or not? :P
    Last edited by metlx; Dec 17th 2009 at 06:39 AM.
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  11. #11
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    Edit* Made some silly errors...
    Last edited by Stroodle; Dec 17th 2009 at 06:27 AM.
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  12. #12
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    Quote Originally Posted by Raoh View Post
    the degree of P is a LOT bigger than $\displaystyle x^2 - 2x + 1$,are you really sure about doing the division.
    that is why i didn't do it

    Quote Originally Posted by Stroodle View Post
    Hi. The way I did it is:

    $\displaystyle p(x)=ax^{2008}+bx^{2007}-1$

    As previously stated, if $\displaystyle (x-1)$ is a factor, then $\displaystyle p(1)=0$

    $\displaystyle \therefore a+b=1$

    Now if you divide $\displaystyle P(x)$ by $\displaystyle (x-1)$ you get:

    $\displaystyle g(x)=ax^{2007}+(b-a)x^{2006}+(a-b)x^{2005}+(b-a)x^{2004}+(a-b)^{2003}.......(b-a)-1$

    Now $\displaystyle g(1)=0$ for $\displaystyle (x-1)$ to be a second factor.

    Which gives $\displaystyle b-1=0$

    $\displaystyle \therefore a=0$ and $\displaystyle b=1$
    i think $\displaystyle g(x) $ should be
    $\displaystyle g(x)=ax^{2007}+(b+a)x^{2006}+(a+b)x^{2005}+(b+a)x^ {2004}+(a+b)x^{2003}+ \cdot \cdot \cdot + (b+a)$

    and

    $\displaystyle g(1) = a + (b+a) + (a+b) + (b+a) + (a+b) + \cdot \cdot \cdot + (b+a) = 0$

    or

    $\displaystyle 2008a + 2007b = 0$

    now, we have two equations in $\displaystyle a$ and $\displaystyle b$, solve it
    Last edited by dedust; Dec 17th 2009 at 03:32 PM. Reason: made a mistake in using horner's rule
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  13. #13
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    im lost.. from that i get

    b = 2007
    a = -2006

    im getting more and more confused.. what is wrong with my solution btw? :P
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  14. #14
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    Quote Originally Posted by metlx View Post

    $\displaystyle x_{1,2} = \frac{-b \pm \sqrt{b^2 -(- 4(1 - b))}}{2(1 - b)} = \frac{-b \pm \sqrt{b^2 - 4b + 4}}{2 - 2b}$
    this formula only work for 2 degrees polynomial
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  15. #15
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    Quote Originally Posted by metlx View Post
    Find a and b (from reals) so that

    $\displaystyle p(x) = ax^{2008} + bx^{2007} - 1$
    will be dividable by polynomial $\displaystyle (x - 1)^2$.

    no idea how to begin.. can someone help?
    HI

    In other words , $\displaystyle (x-1)^2$ would be a factor of $\displaystyle p(x)$ so $\displaystyle p(1)=0$

    $\displaystyle p(1)=a(1)^{2008}+b(1)^{2007}-1$

    $\displaystyle a+b=1$ --- 1

    $\displaystyle p'(x)=2008ax^{2007}+2007bx^{2006}$

    $\displaystyle p'(1)=0$

    $\displaystyle 2008a+2007b=0$ --- 2

    Solving the system would give you $\displaystyle a=-2007$ and $\displaystyle b=2008$
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