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college algebra

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find the complete zeros of the polynomial function. write f in factored form.
f(x)=x^3-7x^2+25x-39
please show all work. I do not understand this.

  • college algebra - ,

    we would let f(x)=x^3-7x^2+25x-39 = 0.

    we would factor to get (x-3)(x^ 2-4x+13).

    find the solutions to the second part (x^2-4x+13)using the quadratic formula. (the first part is x=3)

  • college algebra - ,

    i am not understanding why you put 0 for the first answer. it doesn't seem like you added or did any type of math to get the 0. can you explain more. this is all confusing to me.

  • college algebra - ,

    F(x) = Y = x^3 - 7x^2 + 25x - 39 = 0.
    The real solutions to the Eq are the x-intercepts or the points where the graph crosses the x-axis. The value of Y
    where the curve crosses the x-axis is 0.
    That is why Y is set to 0.

    All real values of X that satisfy the Eq
    is a real solution. It was found by trial and error that 3 gives a zero output(3,0).

    X = 3
    x-3 = 0 and it is a factor.
    Using long-hand division, we divide the
    Eq by x-3 and get x^2-4x+13. Now we have(x-3)(x^2-4x+13) = 0

    x = 4 +-sqrt(16-52)/2
    X = 4 +-sqrt(-36)/2
    x = (4 +-6i)/2 = 2 +- 3i

    Solution set:X = 3,(2+3i), and(2-3i).

  • college algebra - ,

    NOTE: There is only 1 zero(x=3),the
    final factored form is (x-3)(x^2-4x+13)=0.

    The solution set was not required.

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