Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

math problem 3?

Two of the roots of a cubic polynomial with real coefficients are 1+i and 2. If P(1)+P(3)=8, determine the value of P(−1).

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  • MyRank
    Lv 6
    3 weeks ago

    1+i , 1-i  and 2 -roots

    p(x) = (x-α) (x-β) (x-γ)

    p(x) = (x-(1+i))(x-(1-i)) (x-2)

    put x = 1

    p(1) = (1-(1+i))(1-(1-i)) (1-2)

    p(x) = (x-(1+i))(x-(1-i)) (x-2)

    put x = 3

    p(3) = (3-(1+i))(3-(1-i)) (3-2)

    put x = -1

    p(3) = (-1-(1+i))(-1-(1-i)) (-1-2)

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  • Philip
    Lv 6
    4 weeks ago

    P(x) = A(x-2)[(x-1)^2 +1]. In a function with real

    coefficients all complex roots will occur in

    conjugate pairs.

    P(1) = A(-1)1 =-A.

    P(3) = A(5).

    P(1) + P(3) = 4A = 8. Then A = 2.

    Then P(-1) = 2(-3)[(-2)^2 + 1] = -30.

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  • atsuo
    Lv 6
    4 weeks ago

    I think a polynomial has no root , it has zero(s) . 

     

    The third zero is the complex conjugate of 1+i , so it is 1-i . 

    Therefore  the polynomial is 

    A * (x-2) * (x-(1+i)) * (x-(1-i)) 

    = A * (x-2) * ((x-1)+i)) * ((x-1)-i) 

    = A * (x-2) * ((x-1)^2-i^2) 

    = A * (x-2) * (x^2-2x+1+1) 

    = A * (x-2) * (x^2-2x+2) 

    = A * (x^3-4x^2+6x-4) 

     

    P(1) = A * (1-4+6-4) = -A 

    P(3) = A * (27-36+18-4) = 5A 

    P(1)+P(3) = 4A = 8 , so A = 2 . 

     

    The polynomial is 

    2 * (x^3-4x^2+6x-4) = 2x^3-8x^2+12x-8

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