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2 December, 10:28

What are all of the real roots of the following polynomial? f (x) = x5 + 5x4 - 5x3 - 25x2 + 4x + 20

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  1. 2 December, 10:55
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    F (x) = x^5 + 5*x^4 - 5*x^3 - 25*x^2 + 4*x + 20

    By examining the coefficients of the polynomial, we find that

    1+5-5-25+4+20=0 = > (x-1) is a factor

    Now, reverse the sign of coefficients of odd powers,

    -1+5+5-25-4+20=0 = > (x+1) is a factor

    By the rational roots theorem, we can continue to try x=2, or factor x-2=0

    2^5+5 (2^4) - 5 (2^3) - 25 (2^2) + 4 (2) + 20=0

    and similarly f (-2) = 0

    So we have found four of the 5 real roots.

    The remainder can be found by synthetic division as x=-5

    Answer: The real roots of the given polynomial are: {-5,-2,-1.1.2}
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