Find all zeros of F(x) = x^3 - x^2 - x - 2.?

Hi, I am having trouble solving this problem I am not sure if I am supposed to solve it graphically or a different way. Thanks for any help

Find all zeros of F(x) = x^3 - x^2 - x - 2.

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  • 1 decade ago
    Favorite Answer

    x=2,8-4-2-2=0So,(x-2)(x^2+x+1)=0, x=2,x=(-1+3i)/2

    & x=(-1-3i)/2.

    Where i=(-1)^1/2

  • Anonymous
    1 decade ago

    Well, graphing is usually fine but graphing doesn't show irrational roots very nicely. So we have to resort to p/q and then using synthetic division.

    Here p/q or ±2/1 gives us +2 and -2 to check.

    Checking +2: (2)^3-(2)^2-2-2=0

    0=0

    So x=2, or (x-2) is in fact a rational root.

    Now using x-2 and dividing into x^3-x^2-x-2 reveals the quadratic

    x^2+x+1

    And this does indeed have irrational roots that graphing would not have easily revealed.

    Using the quadractic formula on this we find

    x = -1/2±1/2√(3)i

    So there's the three roots or zeros of the cube function, one real and two immaginary.

    Have a good day!

  • Amit Y
    Lv 5
    1 decade ago

    Trying x=2 (a divisor of 2)

    F(2) = 2^3 - 2^2 - 2 - 2 = 8 - 4 - 2 - 2 = 0

    F(x) = x^3 - x^2 - x - 2 = x^3 - 2x^2 + x^2 - 2x + x - 2 =

    = x^2 (x - 2) + x(x - 2) + (x - 2) = (x - 2)(x^2 + x + 1)

    x^2 + x + 1 cannot be further factored in R, because its discriminant

    (-1)^2 - 4*1*1 = 1 - 4 = -3 < 0

    Thus, the only zero is 2.

  • Anonymous
    1 decade ago

    rational zeros?

    +/-2 or +/-1

    try 2:

    8-4-2-2 = 0 , so 2 is a zero, or (x-2) is a factor

    do synthetic division to get it reduced to a quadratic

    2 L 1 -1 -1 -2

    .........2....2...2

    ......1....1....1.0

    F(x) = x^3 - x^2 - x - 2 = (x-2) ( x^2+x+1)

    imaginary solutions?? depends on your level?

    if only real solutions: x = 2

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