math

find the real zeros of the given functions, use the real zeros to factor f. f(x)=-2x4power+9x3power+2x2power-39x+18

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  1. I found f(3) = 0 , x-3 is a factor
    then found f(-2) = 0 , so x+2 is a factor
    divided both of those and got
    -(2x^2-7x+3)
    which factored further to -(x-3)(2x-1)

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

    with zeros at x= -2, 1/2 and 3 , 3 being a double root

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  2. 234357

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  3. y = -2 x^4 + 9 x^3 + 2 x^2 - 39 x + 18

    make a table of x and y
    x _____ y

    -2 _____ 0 oh wow

    -1 _____ 48

    0 ______ 18

    +1 ______ -12 so a zero between 0 and 1

    well, at least we had a zero at x = -2 so (x+2) is a factor
    so divide this mess by (x+2)
    so I have
    (x+2)(-2x^3+13x^2-24x+9)
    well, try .5 or divide by (2x-1)
    because we have that 2 there and we know we have a zero around .5
    It works and I have
    (x+2)(2x-1)(-x^2+6x-9)
    well 9 does not factor by much so try 3
    -(3)^2 + 6(3) -9 = 18 -18 = 0
    so I have
    (x+2)(2x-1)(x-3)(x-3) but we want negative those (zero has either sign) so multiply by -1
    (-1)(x+2)(2x-1)(x-3)(x-3)

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