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Thread: Help with functions or whatever it's name is

  1. #1
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    Help with functions or whatever it's name is

    Hello guys, i've got this report i have to do and would really need your help.. This will propably be a piece of cake for you guys. Here it is:

    Which value for the constant a is the polynom Q(x)=2ax^3-x+a^2 divideable with
    a) x+1 b)2x+1
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  2. #2
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    Quote Originally Posted by FelluRan View Post
    Hello guys, i've got this report i have to do and would really need your help.. This will propably be a piece of cake for you guys. Here it is:

    Which value for the constant a is the polynom Q(x)=2ax^3-x+a^2 divideable with
    a) x+1 b)2x+1
    Use the factor theorem: If $\displaystyle (x-p)$ is a factor of $\displaystyle f(x) $then $\displaystyle f(p) = 0$

    $\displaystyle Q(-1) = 2a(-1)^3-(-1)+a^2 = a^2-2a+1 = (a-1)^2 = 0$

    $\displaystyle \therefore a = 1$
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    Could you please write more specificly what you have done ? (I know i'm stupid )
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    Factor theorem - Wikipedia, the free encyclopedia

    $\displaystyle Q(x)=2ax^3-x+a^2$

    The factor theorem says that $\displaystyle (x-p)$ is a factor of $\displaystyle f(x)$ only if $\displaystyle f(p) = 0$

    As the question says find the value of $\displaystyle a$ for which $\displaystyle x+1$ is a factor of $\displaystyle Q(x)$ this means the factor theorem is applied and so $\displaystyle Q(-1) = 0$

    This means we find the value of Q(-1) by substituting $\displaystyle -1 $ wherever $\displaystyle x$ is seen and setting $\displaystyle Q(x)$ to $\displaystyle 0$

    $\displaystyle Q(-1) = 2a(-1)^3 -(-1) + a^2 = 0$

    This simplifies to $\displaystyle a^2-2a+1=0$ which in turn simplifies to $\displaystyle (a-1)^2 = 0$ and hence it's clear that a = 1
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    To be precise, $\displaystyle a = 1$ found in this way is the necessary, but not automatically sufficient, condition for $\displaystyle Q(a, x)$ to be divisible by $\displaystyle x + 1$. Therefore, to show that it is sufficient, one can actually verify that $\displaystyle x + 1$ divides $\displaystyle Q(1, x)$ with no remainder.
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