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Thread: Find function from remainders

  1. #1
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    Find function from remainders

    Determine $\displaystyle g(x)\equiv (px+q)^2(2x^2-5x-3)$ where p,q are constants, given that, on division by $\displaystyle (x-2)^2$, the remainder is $\displaystyle -39-3x$.

    I found
    $\displaystyle g(2)=(2p+q)^2(-5)=-45$
    $\displaystyle (2p+q)^2=9$
    $\displaystyle 2p+q=\pm 3$

    I don't think I'm on the right track, but I don't know how else to work it out.
    Thanks!
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  2. #2
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    Good start. I am not sure if the following is the simplest way, but it seems to lead to finding p and q, at least in theory.

    So, we have $\displaystyle (px+q)^2(2x^2-5x-3)=(x-2)^2f(x)-39-3x$ for some function $\displaystyle f(x)$. Now, looking at degrees of polynomials, $\displaystyle f(x)$ is a quadratic polynomial, let's say $\displaystyle ax^2+bx+c$.

    Next, we look at both sides' coefficients of $\displaystyle x^4$, $\displaystyle x^3$, as well as the free terms. Equating the leading coefficients, we see that $\displaystyle a=2p^2$. Similarly, equating the constant terms (or putting x = 0), we get that $\displaystyle c=(39-3q^2)/4$. Finally, equating the coefficients at $\displaystyle x^3$, we get that $\displaystyle b=3p^2+4pq$. Therefore, $\displaystyle f(x)=2p^2+(3p^2+4pq)x+(39-3q^2)/4$.

    Next, one of the roots of $\displaystyle 2x^2-5x-3=0$ is -1/2. Substituting $\displaystyle x=-1/2$ into the original equation, we get $\displaystyle f(-1/2)=-15$, or $\displaystyle 4f(-1/2)=-60$. Substitute the expression for $\displaystyle f(x)$ above; this gives (after lots of simplifications) $\displaystyle 4p^2+3q^2+8pq=99$.

    Finally, combining this with $\displaystyle 2p+q=\pm 3$, one can find $\displaystyle p$ and $\displaystyle q$. In my calculations, this boils down to a linear (not even a quadratic) equation.

    Disclaimer: my calculations can be wrong.
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  3. #3
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    i solved it, but there were 2 possible values for p and 4 possible values for q. which is too many.
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  4. #4
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    In my calculations, it's p = -6 and q = 15 or p = 6 and q = -15. It does not matter which is chosen since px+q is squared.
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  5. #5
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    the answer is p=1, q=-1. what i got was p=1, q=1.
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  6. #6
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    p=1, q=-1 does not fit 2p + q = 3.
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  7. #7
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    Given problem is

    $\displaystyle (px + q)^2(2x^2 - 5x - 3)$

    $\displaystyle (px + q)^2[(2x^2 - 8x +8) +3x -11]$

    Since (2x^2 - 8x +8) is divisible by (x-2)^2

    $\displaystyle (px + q)^2(3x - 11) = (x - 2)^2(ax + b) - 39 - 3x$

    Compare the coefficients of x^3....> 3p^2 = a...(1)

    Compare the coefficients of x^2...> 6pq - 11p^2 = -4a + b....(2)

    Compare the coefficients of x....> 3q^2 - 22q = 4a - 4b - 3....(3)

    Compare the constants....> -11q^2 = 4b - 39...(4)

    Adding them you get

    -8(p+q)^2 = a + b - 42

    To satisfy equation, if p+q = 2, a+b should be 10.

    from eq(1) if p = 1 , a = 3.

    from eq.(2) if q = 1 , b = 7.

    Hence p = 1 and q = 1.
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  8. #8
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    This is the answer given: $\displaystyle (x-1)^2(2x^2-5x-3)$
    So it means that p=1, and q=-1.
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  9. #9
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    You can cross check by dividing the given polynomial by ( x-2)^2 and finding the reminder.
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  10. #10
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    Quote Originally Posted by sa-ri-ga-ma View Post

    To satisfy equation, if p+q = 2, a+b should be 10.
    How do I know p+q=2?
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