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Math Help - values of roots

  1. #1
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    values of roots

    hi, the question i've got is:

    the polynomial P(x) = x^3 - 4x^2 + 2x + 4 has (x - 2) as a factor

    find the exact values of the roots of the equation

    could someone show me step by step how you would arrive at the answer please? thankyou
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  2. #2
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    Quote Originally Posted by mark View Post
    hi, the question i've got is:

    the polynomial P(x) = x^3 - 4x^2 + 2x + 4 has (x - 2) as a factor

    find the exact values of the roots of the equation

    could someone show me step by step how you would arrive at the answer please? thankyou
    P(x) = (x-2)(ax^2 + bx + c)

    Where a,b and c are constants. You can find a, b and c by comparing coefficients

    Spoiler:
    .These will be of the form P(x) = (x-2)(ax^2+bx+c)

    x^3 : 1 = a

    x^2 : -4 = -2a+b \rightarrow b = -2

    x^0 (constant) : 4 = -2c \rightarrow c = -2

    Therefore P(x) = (x-2)(x^2-2x-2)
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  3. #3
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    i already did that bit, the answer in the book is x =2, x = 1 + \sqrt{3} and x = 1 - \sqrt{3}

    any ideas on how to arrive at that answer?
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  4. #4
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    Quote Originally Posted by mark View Post
    i already did that bit, the answer in the book is x =2, x = 1 + \sqrt{3} and x = 1 - \sqrt{3}

    any ideas on how to arrive at that answer?
    The roots are where it equals 0 at.

    (x-2)(x^2-2x-2) = 0

    So either x-2=0 or x^2-2x-2=0

    The book's answers are the solutions to those equations
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  5. #5
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    ok thanks, how would you get to x = 1 + \sqrt{3} and x = 1 - \sqrt{3} from (x^2 - 2x -2) though? what formula would you use?
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  6. #6
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    Quote Originally Posted by mark View Post
    ok thanks, how would you get to x = 1 + \sqrt{3} and x = 1 - \sqrt{3} from (x^2 - 2x -2) though? what formula would you use?
    The quadratic formula:

    For ax^2+bx+c=0:

    x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}
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  7. #7
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    ahhhh the quadratic formula, thanks for the help
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