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Thread: more constructing polynomials.

  1. #1
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    more constructing polynomials.

    If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
    cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""

    AND

    Show that the equation $\displaystyle x^4+x^3+x+1=0$ has a double root at x=-1. Hence show that there are no other real roots.
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    Quote Originally Posted by THSKluv View Post
    If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
    cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""

    AND

    Show that the equation $\displaystyle x^4+x^3+x+1=0$ has a double root at x=-1. Hence show that there are no other real roots.
    1. For the polynomial $\displaystyle P(x) = 2x^4 + 8x^3 + px^2 + qx + r$, if $\displaystyle x = -1, x = 1, x = 3$ are roots of the polynomial, then $\displaystyle (x +1), (x - 1), (x - 3)$ are factors of the polynomial.

    So that means, by the factor and remainder theorems...

    $\displaystyle P(-1) = 0, P(1) = 0, P(3) = 0$.


    So...
    $\displaystyle 2(-1)^4 + 8(-1)^3 + p(-1)^2 + q(-1) + r = 0$

    $\displaystyle 2 - 8 + p - q + r = 0$

    $\displaystyle -6 + p - q + r = 0$

    $\displaystyle p - q + r = 6$.



    $\displaystyle 2(1)^4 + 8(1)^3 + p(1)^2 + q(1) + r = 0$

    $\displaystyle 2 + 8 + p + q + r = 0$

    $\displaystyle 10 + p + q + r = 0$

    $\displaystyle p + q + r = -10$.



    $\displaystyle 2(3)^4 + 8(3)^3 + p(3)^2 + q(3) + r = 0$

    $\displaystyle 162 + 216 + 9p + 3q + r = 0$

    $\displaystyle 378 + 9p + 3q + r = 0$

    $\displaystyle 9p + 3q + r = -378$.



    Now you have three equations in three unknowns which you can solve simultaneously for $\displaystyle p, q, r$.

    $\displaystyle p - q + r = 6$

    $\displaystyle p + q + r = -10$

    $\displaystyle 9p + 3q + r = -378$.
    Last edited by Prove It; Jul 3rd 2009 at 11:19 PM.
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  3. #3
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    Quote Originally Posted by THSKluv View Post
    If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
    cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""

    AND

    Show that the equation $\displaystyle x^4+x^3+x+1=0$ has a double root at x=-1. Hence show that there are no other real roots.
    2. To show that $\displaystyle x^4 + x^3 + x + 1 = 0$ has a double root at $\displaystyle x = -1$, you have to show that $\displaystyle (x + 1)^2$ is a factor.


    $\displaystyle x^4 + x^3 + x + 1 = 0$

    $\displaystyle x^3(x + 1) + 1(x + 1) = 0$

    $\displaystyle (x + 1)(x^3 + 1) = 0$

    $\displaystyle (x + 1)(x + 1)(x^2 - x + 1) = 0$

    $\displaystyle (x + 1)^2(x^2 - x + 1) = 0$.
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    1) If -1, 1, 3 are the roots then

    $\displaystyle \left\{\begin{array}{lll}P(-1)=0\\P(1)=0\\P(3)=0\end{array}\right.\Rightarrow\ left\{\begin{array}{lll}p-q+r=6\\p+q+r=-10\\9p+3q+r=-378\end{array}\right.$

    Substracting the first equation from the second we have $\displaystyle 2q=-16\Rightarrow q=-8$

    Replacing q in the first and the third equation we have

    $\displaystyle \left\{\begin{array}{ll}p+r=-2\\9p+r=-354\end{array}\right.$

    Substracting the first from the second we get $\displaystyle 8p=-352\Rightarrow p=-44\Rightarrow r=42$

    2) $\displaystyle x^4+x^3+x+1=c^3(x+1)+x+1)=(x+1)(x^3+1)=(x+1)^2(x^2-x+1)$

    Then $\displaystyle x_1=x_2=-1$ and $\displaystyle x^2-x+1=0$ has no real roots.
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    Quote Originally Posted by red_dog View Post
    1) If -1, 1, 3 are the roots then

    $\displaystyle \left\{\begin{array}{lll}P(-1)=0\\P(1)=0\\P(3)=0\end{array}\right.\Rightarrow\ left\{\begin{array}{lll}p-q+r=6\\p+q+r=-10\\9p+3q+r=-378\end{array}\right.$

    Substracting the first equation from the second we have $\displaystyle 2q=-16\Rightarrow q=-8$

    Replacing q in the first and the third equation we have

    $\displaystyle \left\{\begin{array}{ll}p+r=-2\\9p+r=-354\end{array}\right.$

    Substracting the first from the second we get $\displaystyle 8p=-352\Rightarrow p=-44\Rightarrow r=42$

    2) $\displaystyle x^4+x^3+x+1=c^3(x+1)+x+1)=(x+1)(x^3+1)=(x+1)^2(x^2-x+1)$

    Then $\displaystyle x_1=x_2=-1$ and $\displaystyle x^2-x+1=0$ has no real roots.
    How do you get $\displaystyle 9p + 3q + r = -378$?
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  6. #6
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Prove It View Post
    How do you get $\displaystyle 9p + 3q + r = -378$?
    Replace $\displaystyle x=3$:

    $\displaystyle 2\cdot3^4+8\cdot 3^3+p\cdot3^2+q\cdot 3+r=0$

    $\displaystyle 162+216+9p+3q+r=0\Rightarrow 9p+3q+r=-378$
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  7. #7
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    Quote Originally Posted by red_dog View Post
    Replace $\displaystyle x=3$:

    $\displaystyle 2\cdot3^4+8\cdot 3^3+p\cdot3^2+q\cdot 3+r=0$

    $\displaystyle 162+216+9p+3q+r=0\Rightarrow 9p+3q+r=-378$
    Thanks. I had a small error - my arithmetic is well-known to be shocking.
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  8. #8
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    Quote Originally Posted by THSKluv View Post
    If the roots of $\displaystyle 2x^4+8x^3+px^2+qx+r=0$ are -1, 1, 3, find the values of p, q,r
    cant seem to get the right answer, i thought the last root must be 1 because the sum of the roots must be 4 (8/2) cept the answers say that p=-44, q=-8, r=42 which doesnt work for me ><""
    The sum of the roots is -4 not 4, so the missing root is -7.

    CB
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