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  1. #1
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    polynomial division problem

    Find all values of k such that f(x) = kx^3 + x^2 + k^2x + 3k2 + 11 is completely divisible by polynomial p(x) = x + 2


    I have no idea on how to solve this.
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  2. #2
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    Quote Originally Posted by mwok View Post
    Find all values of k such that f(x) = kx^3 + x^2 + k^2x + 3k2 + 11 is completely divisible by polynomial p(x) = x + 2


    I have no idea on how to solve this.
    Well, mwok. Let's try synthetic division on this one.

    $\displaystyle f(x)=kx^3+x^2+k^2x+3k^2+11$. Since $\displaystyle (x + 2)$ is a factor, then $\displaystyle f(-2)=0$.


    Code:
     
    -2 | k      1         k^2      3k^2  +11
              -2k        4k-2     -2k^2-8k+4
        ------------------------------------
         k    -2k+1    k^2+4k-2     k^2-8k+15
    Now then, we know the remainder should be 0, so....

    $\displaystyle k^2-8k+15=0$

    Solve for the two possible values of k and you're done.
    Last edited by masters; Oct 29th 2008 at 12:38 PM.
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    Quote Originally Posted by masters View Post
    Well, mwok. Let's try synthetic division on this one.

    $\displaystyle kx^3+x^2+k^2x+3k^2+11$. Since $\displaystyle (x + 2)$ is a factor, then $\displaystyle f(-2)=0$.


    Code:
     
    -2 | k      1         k^2      3k^2  +11
              -2k        4k-2     -2k^2-8k+4
        ------------------------------------
         k    -2k+1    k^2+4k-2     k^2-8k+15
    Now then, we know the remainder should be 0, so....

    $\displaystyle k^2-8k+15=0$

    Solve for the two possible values of k and you're done.

    I'm not sure I understand, with synthetic divisions...don't you just use the coefficients?
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  4. #4
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    Quote Originally Posted by mwok View Post
    I'm not sure I understand, with synthetic divisions...don't you just use the coefficients?
    I did use the coefficients.

    $\displaystyle f(x)={\color{red}k}x^3+{\color{red}1}x^2+{\color{r ed}k^2}x+{\color{blue}3k^2+11}$.

    The coefficients are in red and the constant is in blue.
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  5. #5
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    x is the variable in the polynomial, not k. k is an undetermined coefficient, but it is still a coefficient, so you can use synthetic division and get an equation for k, and figure out what value it has.
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    Quote Originally Posted by masters View Post
    I did use the coefficients.

    $\displaystyle f(x)={\color{red}k}x^3+{\color{red}1}x^2+{\color{r ed}k^2}x+{\color{blue}3k^2+11}$.

    The coefficients are in red and the constant is in blue.

    Sorry, I'm a bit confused.

    Why do you use a constant in synthetic division? I thought synthetic division only used the coefficients.


    Attached is what I got.
    Attached Thumbnails Attached Thumbnails polynomial division problem-math.jpg  
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    Quote Originally Posted by mwok View Post
    Sorry, I'm a bit confused.

    Why do you use a constant in synthetic division? I thought synthetic division only used the coefficients.


    Attached is what I got.
    There are only 3 terms with an x in them. In those 3 terms, whatever has been multiplied by x is a coefficient. Any other terms (without x) would constitute the constant.

    $\displaystyle f(x)=ax^2+bx+c$

    a is the coefficient of the quadratic term.
    b is the coefficient of the linear term.
    c is the constant.

    It just so happens that the c in your function is $\displaystyle 3k^2+11$, but a constant none the less.

    Quote Originally Posted by mwok View Post
    I thought synthetic division only used the coefficients.
    Not so. Your attachment is incorrect. Review Synthetic Division
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    As was demonstrated in my post #2, k can be found by solving the following equation:

    $\displaystyle k^2-8k+15=0$

    $\displaystyle (k-5)(k-3)=0$

    $\displaystyle k=5 \ \ or \ \ k=3$

    If you substitute each of these values for k back into your original function, you can verify that f(-2)=0.
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  9. #9
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    Hello, mwok

    Find all values of $\displaystyle k$ such that $\displaystyle f(x) \:= \:kx^3 + x^2 + k^2x + 3k^2 + 11$ is divisible by $\displaystyle x + 2$
    We are expected to know this theorem:

    . . If $\displaystyle f(a) \:=\:0$, then $\displaystyle (x-a)$ is a factor of $\displaystyle f(x)$.


    For this problem, if $\displaystyle f(\text{-}2) = 0$, then $\displaystyle (x + 2)$ is a factor of $\displaystyle f(x).$

    We have: .$\displaystyle f(\text{-}2) \;=\;k(\text{-}2)^4 + (\text{-}2)^2 + k^2(\text{-}2) + 3k^2 + 11 \;=\;0$

    . . which simplifies to: .$\displaystyle k^2-8k+15 \;=\;0$

    . . which factors: .$\displaystyle (k-3)(k-5) \:=\:0$

    . . and has roots: .$\displaystyle \boxed{k \;=\;3,\:5}$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    If you dare, you can try Long Division . . .



    $\displaystyle \begin{array}{ccccccc} & & & kx^2 & +(1-2k)x & +(k^2+4k -2) \\
    & & ---&-----&-------&-------&--- \\
    x+2 & ) & kx^3 & +x^2 & +k^2x & + 3k^2 & +11 \\
    & & kx^3 & +2kx^2 \\
    & & --- & ----- \\
    & & & (1-2k)x^2 & +k^2x \\
    & & & (1-2k)x^2 & + 2(1-2k)x \\
    & & & ----- & ------- \\
    & & & & (k^2+4k-1)x & + 3k^2 & +11\\
    & & & & (k^2+4k-2)x & +2(k^2+4k-2) \\
    & & & & -------& ------- \\\end{array}$

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .$\displaystyle k^2 - 8k + 4 \quad\;\;+ 11$


    If the division "comes out even", then the remainder is zero.

    So we have: .$\displaystyle k^2 - 8k + 15 \:=\:0 \quad\hdots\quad See?$

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  10. #10
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    Thanks for the help! Will read on synthetic division further.
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