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Math Help - Polynomial coefficient problem

  1. #1
    Junior Member PaulAdrienMauriceDirac's Avatar
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    Polynomial coefficient problem

    Hello, I have an another unsolved problem from Gelfand's book on Algebra, here it is:
    Problem 153. The polynomial P=x^3+x^2-10x+1 has three different roots (the authors guarantee it) denoted by x_1,x_2,x_3. Write a polynomial with integer coefficients having roots
    (a) x_1+1, x_2+1, x_3+1; (b) 2x_1, 2x_2, 2x_3; (c) \frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}

    I understand that P=x^3+x^2-10x+1=(x-x_1)(x-x_2)(x-x_3), but that's pretty much all I can 'do', hints would be appreciated, thanks in advance.
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    Re: Polynomial coefficient problem

    Hello, PaulAdrienMauriceDirac!

    I'll do part (a).
    I hope you understand the reasoning.


    \text{Polynomial }P(x)\:=\:x^3+x^2-10x+1\,\text{ has three roots, denoted }a,\,b,\,c.

    \text{Write a polynomial }Q(x)\text{ with integer coefficients having roots:}

    (a)\;a+1,\:b+1,\:c+1 \qquad (b)\;2a,\,2b,\,2c \qquad (c)\;\frac{1}{a},\:\frac{1}{b},\:\frac{1}{c}

    We know that: . \begin{Bmatrix} a+b+c \:=\:\text{-}1 \\ ab+bc+ac \:=\:\text{-}10 \\ abc \:=\:\text{-}1 \end{Bmatrix}


    (a)\text{ Roots: }\:a+1,\:b+1,\:c+1


    Sum of roots: . (a+1)+(b+1)+(c+1) \;=\;(a+b+c) + 3 \;=\;\text{-}1 + 3 \;=\;2


    Sum of pairs: . (a+1)(b+1) + (b+1)(c+1) + (a+1)(c+1)

    . . . . . . . . . =\;(ab + bc + ac) + 2(a+b+c) + 1

    . . . . . . . . . =\;(\text{-}10) + 2(\text{-}1) + 3\;=\;\text{-}9


    Product of roots: . (a+1)(b+1)(c+1)

    . . . . . . . . . . . =\;abc + (ab + bc + ac) + (a+b+c) + 1

    . . . . . . . . . . . =\;\text{-}1 + (\text{-}10) + (\text{-}1) + 1 \;=\;\text{-}11


    Therefore: . Q(x) \;=\;x^3 - 2x^2 - 9x + 11
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial coefficient problem

    Quote Originally Posted by Soroban View Post
    Hello, PaulAdrienMauriceDirac!

    I'll do part (a).
    I hope you understand the reasoning.



    We know that: . \begin{Bmatrix} a+b+c \:=\:\text{-}1 \\ ab+bc+ac \:=\:\text{-}10 \\ abc \:=\:\text{-}1 \end{Bmatrix}


    (a)\text{ Roots: }\:a+1,\:b+1,\:c+1


    Sum of roots: . (a+1)+(b+1)+(c+1) \;=\;(a+b+c) + 3 \;=\;\text{-}1 + 3 \;=\;2


    Sum of pairs: . (a+1)(b+1) + (b+1)(c+1) + (a+1)(c+1)

    . . . . . . . . . =\;(ab + bc + ac) + 2(a+b+c) + 1

    . . . . . . . . . =\;(\text{-}10) + 2(\text{-}1) + 3\;=\;\text{-}9


    Product of roots: . (a+1)(b+1)(c+1)

    . . . . . . . . . . . =\;abc + (ab + bc + ac) + (a+b+c) + 1

    . . . . . . . . . . . =\;\text{-}1 + (\text{-}10) + (\text{-}1) + 1 \;=\;\text{-}11


    Therefore: . Q(x) \;=\;x^3 - 2x^2 - 9x + 11
    Thank you very much! I didn't know that Vieta's theorem existed for cubic equations, your explanation is very clear.
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    Re: Polynomial coefficient problem

    Hi PaulAdrienMauriceDirac,

    P(x) = (x - x1)(x - x2)(x - x3) = x^3 - (x1 + x2 + x3)x^2 + (x1x2 + x1x3 + x2x3)x - x1x2x3.
    So, equating co-efficients like this holds for cubics, and in general, all polynomial equations roots.

    Salahuddin
    Maths online
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    Re: Polynomial coefficient problem

    While Vieta's theorem is the way to go for parts b) and c), for part a) you could simply shift the polynomial 1 unit to the right to add 1 to the roots:

    f(x)=(x-1)^3+(x-1)^2-10(x-1)+1=

    x^3-3x^2+3x-1+x^2-2x+1-10x+10+1=

    x^3-2x^2-9x+11
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  6. #6
    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial coefficient problem

    Quote Originally Posted by MarkFL2 View Post
    While Vieta's theorem is the way to go for parts b) and c), for part a) you could simply shift the polynomial 1 unit to the right to add 1 to the roots:

    f(x)=(x-1)^3+(x-1)^2-10(x-1)+1=

    x^3-3x^2+3x-1+x^2-2x+1-10x+10+1=

    x^3-2x^2-9x+11
    Oh, that's so much easier, too bad I haven't thought about it, thank you very much!
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    Re: Polynomial coefficient problem

    The same sort of method will work for part (b) as well, replace x by x/2.
    The resulting polynomial is
    \frac{x^{3}}{8}+\frac{x^{2}}{4}-5x+1.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial coefficient problem

    Quote Originally Posted by BobP View Post
    The same sort of method will work for part (b) as well, replace x by x/2.
    The resulting polynomial is
    \frac{x^{3}}{8}+\frac{x^{2}}{4}-5x+1.
    Hm, are you sure that that method works on part (b) as well? because I got different results when using Vieta's theorem.

    P(x)=x^3+px^2+qx+r
    We know from Vieta's that:
    x_1+x_2+x_3=-p;

    x_1x_2+x_1x_3+x_2x_3=q;

    x_1x_2x_3=-r

    So:
    x_1+x_2+x_3=-1 -> 2x_1+2x_2+2x_3=2(x_1+x_2+x_3)=2(-1)=-2
    x_1x_2+x_1x_3+x_2x_3=-10 -> 4x_1x_2+4x_1x_3+4x_2x_3=4(x_1x_2+x_1x_3+x_2x_3)=-40
    x_1x_2x_3=-1 -> 2x_1*2x_2*2x_3=8(x_1x_2x_3)=8(-1)=-8

    Therefore:
    p=2, q=-40, r=8 and Q(x)=x^3+2x^2-40x+8
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  9. #9
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    Re: Polynomial coefficient problem

    Your polynomial is simply eight times mine, they will have the same zeros.
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