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Math Help - Polynomial divison.

  1. #1
    ohm
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    Polynomial divison.

    if x^3+px+7 is completely divisible by qx^2+r, then find the relation between p,q and r.
    I have tried to solve it by putting the value of x^2=-r/q but unable to reach any solution.
    Last edited by mr fantastic; July 22nd 2010 at 05:27 AM. Reason: Re-titled.
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  2. #2
    Senior Member Dinkydoe's Avatar
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    Are you sure this is the problem? Unless q=0 it seems to me impossible
    Last edited by Dinkydoe; July 21st 2010 at 09:50 AM.
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    What makes you say that, Dinky? There's limitless solutions.
    Example: x=2,p=18,q=3,r=5
    x^3 + px + 7 = 51
    qx^2 + r = 17
    51/17 = 3
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  4. #4
    Senior Member Dinkydoe's Avatar
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    That's true however, I'm not so sure how to interpret the problem.

    If we just see it as polynomial division (and don't give x a specific value)

    Then it's just like I said.

    x^3+px+7 = (qx^2+r)(ax+b) = aqx^3+bqx^2+arx+br \Rightarrow

    bq = 0 but b= 7/r\neq 0

    We get q= 0
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  5. #5
    ohm
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    Quote Originally Posted by Wilmer View Post
    What makes you say that, Dinky? There's limitless solutions.
    Example: x=2,p=18,q=3,r=5
    x^3 + px + 7 = 51
    qx^2 + r = 17
    51/17 = 3
    Sorrư! How have you reached this solution? And what is the relation between p,q and r?
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  6. #6
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    Hello, ohm!

    if x^3+px+7 is divisible by qx^2+r,
    then find the relation between p,q and r.

    This is a Factoring problem.
    The final result should be true for all values of x.
    . . We are not allowed to insert constant values for x.


    Otherwise, this example would be correct.

    . . Is x^2 + 9 divisible by x+1 ?

    . . Answer: .Yes, let x = 4.


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


    I agree with Dinkydoe . . . The problem is faulty.


    Since  x^3 + px + 7 is divisible by qx^2 + r,

    . . we have: . (ax+b)(qx^2+r) \:=\:x^2 + px + 7

    . . . . for some binimial, ax+b.


    Hence: . aqx^3 + bqx^2 + arx + br \;=\;x^3 + px + 7


    Equate coefficients: . \begin{Bmatrix}aq &=& 1 & [1] \\<br />
bq &=& 0 & [2] \\ ar &=& p & [3] \\ br &=& 7 & [4] \end{Bmatrix}


    From [4], we have: . b \ne 0,\;r \ne 0

    In [2] we have: . bq = 0,\;b \ne 0 \quad\Rightarrow\quad q = 0

    In [1] we have: . aq = 1,\;q = 0 \quad\Rightarrow\quad a\!\cdot\!0 = 1 .??

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  7. #7
    Member Mathelogician's Avatar
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    I think you shuld solve it thus;
    Let A(x)=x^3+px+7.
    By Polynomial long division we know that A(x)=(qx^2+r)Q(x)+R(x) and To be A(x) divisible by qx^2+r, we must have R(x)=0.
    If we divide A(x) by qx^2+r, we earn A(x)= [(x/q)(qx^2+r)]+((pq-r)/q)x+7.
    Therefore by my last assertion we have ((pq-r)/q)x+7=0 => x=-7/(p+(r/q)).
    This means that if x=-7/(p+(r/q)), then R(x)=0,
    So subsituate this value for x in the main division expression:
    A(-7/(p+(r/q)))=q [(-7/(p+(r/q)))^2+r]Q(x).
    Now you can find the relation between p,q and r.
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  8. #8
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    Quote Originally Posted by ohm View Post
    if x^3+px+7 is completely divisible by qx^2+r, then find the relation between p,q and r.
    I have tried to solve it by putting the value of x^2=-r/q but unable to reach any solution.
    If we write x^3+px+7=\left(\frac{x}{q}+\frac{7}{r}\right)\left  (qx^2+r\right)

    then x^3+px+7=x^3+\frac{7q}{r}x^2+\frac{r}{q}x+7

    giving px=\frac{7q}{r}x^2+\frac{r}{q}x

    which imposes a restriction on x rather than allowing x to be the set of reals.

    For example, q=1, r=1... then choosing x=2 and p=15 is a solution
    but not for x in general.

    For all x, we'd require px=\frac{7q}{r}x^2+\frac{r}{q}x

    or q=0\ if\ \frac{r}{q}=p

    Equating coefficients works if we are dealing with x as the set of reals or complex numbers.


    If we perform polynomial division, we get

    \frac{x^3+px+7}{qx^2+r}=\frac{x}{q}+\frac{\left(\f  rac{pq-r}{q}\right)x+7}{qx^2+r}

    If we say that the numerator of the rightmost term must be zero, we impose the following restriction on x...

    x^3+px+7=x^3+\frac{r}{q}x\ \Rightarrow\ px+7=\frac{r}{q}x

    x=1,\ p=-6,\ q=r=2 is one way to satisfy that.

    The restriction \frac{r-pq}{q}x=7 is the above restriction.

    Also, in saying the rightmost numerator must be zero, we get "in general" 7=0 and p=\frac{r}{q} by equating coefficients.
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