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Math Help - more constructing polynomials.

  1. #1
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    more constructing polynomials.

    If the roots of 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 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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  2. #2
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    Quote Originally Posted by THSKluv View Post
    If the roots of 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 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 P(x) = 2x^4 + 8x^3 + px^2 + qx + r, if x = -1, x = 1, x = 3 are roots of the polynomial, then (x +1), (x - 1), (x - 3) are factors of the polynomial.

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

    P(-1) = 0, P(1) = 0, P(3) = 0.


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

    2 - 8 + p - q + r = 0

    -6 + p - q + r = 0

    p - q + r = 6.



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

    2 + 8 + p + q + r = 0

    10 + p + q + r = 0

    p + q + r = -10.



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

    162 + 216 + 9p + 3q + r = 0

    378 + 9p + 3q + r = 0

    9p + 3q + r = -378.



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

    p - q + r = 6

    p + q + r = -10

    9p + 3q + r = -378.
    Last edited by Prove It; July 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 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 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 x^4 + x^3 + x + 1 = 0 has a double root at x = -1, you have to show that (x + 1)^2 is a factor.


    x^4 + x^3 + x + 1 = 0

    x^3(x + 1) + 1(x + 1) = 0

    (x + 1)(x^3 + 1) = 0

    (x + 1)(x + 1)(x^2 - x + 1) = 0

    (x + 1)^2(x^2 - x + 1) = 0.
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  4. #4
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    1) If -1, 1, 3 are the roots then

    \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 2q=-16\Rightarrow q=-8

    Replacing q in the first and the third equation we have

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

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

    2) 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 x_1=x_2=-1 and 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

    \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 2q=-16\Rightarrow q=-8

    Replacing q in the first and the third equation we have

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

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

    2) 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 x_1=x_2=-1 and x^2-x+1=0 has no real roots.
    How do you get 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 9p + 3q + r = -378?
    Replace x=3:

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

    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 x=3:

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

    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 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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