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Math Help - help solve 3

  1. #1
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    help solve 3

    The cubic polynomial P(x) =x3 +rx2 +sx +t
    , where r, s and t are real numbers, has three real zeros, 1, αand α.
    (i) Find the value of r.
    (ii) Find the value of s +t.
    P(x) =x3 +rx2 +sx +t.

    with working out please.
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  2. #2
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    Quote Originally Posted by andy69 View Post
    The cubic polynomial P(x) =x3 +rx2 +sx +t
    , where r, s and t are real numbers, has three real zeros, 1, αand –α.
    (i) Find the value of r.
    (ii) Find the value of s +t.
    P(x) =x3 +rx2 +sx +t.

    with working out please.
    This is not as hard as it may look.

    Write P(x) = (x-1)(x-\alpha)(x-(-\alpha)), expand, and then match the coefficients.

    Note that (x-\alpha)(x+\alpha)=x^2-\alpha^2 (in other words, you can skip the step of writing out that expansion explicitly by noticing there is a difference of squares).
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  3. #3
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    Quote Originally Posted by undefined View Post
    This is not as hard as it may look.

    Write P(x) = (x-1)(x-\alpha)(x-(-\alpha)), expand, and then match the coefficients.

    Note that (x-\alpha)(x+\alpha)=x^2-\alpha^2 (in other words, you can skip the step of writing out that expansion explicitly by noticing there is a difference of squares).
    so after its difference between the squares x^2-a^2 do you expand it with (x-a)
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  4. #4
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    Quote Originally Posted by andy69 View Post
    so after its difference between the squares x^2-a^2 do you expand it with (x-a)
    I mentioned the difference of squares as a "shortcut."

    Normally one might write

    (x-\alpha)(x+\alpha) = x^2 + x\alpha - x\alpha - \alpha^2 = x^2 -\alpha^2.

    Recognising a difference of squares allows you to skip the middle step and instantly jump to the last step.

    Anyway, what you'll want is

    P(x)=(x-1)(x^2-\alpha^2)= x^3 - x^2 -\alpha^2x + \alpha^2

    which allows you to write

    r=-1

    s=-\alpha^2

    t=\alpha^2
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  5. #5
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    Quote Originally Posted by undefined View Post
    I mentioned the difference of squares as a "shortcut."

    Normally one might write

    (x-\alpha)(x+\alpha) = x^2 + x\alpha - x\alpha - \alpha^2 = x^2 -\alpha^2.

    Recognising a difference of squares allows you to skip the middle step and instantly jump to the last step.

    Anyway, what you'll want is

    P(x)=(x-1)(x^2-\alpha^2)= x^3 - x^2 -\alpha^2x + \alpha^2

    which allows you to write

    r=-1

    s=-\alpha^2

    t=\alpha^2
    so answering my questions
    (i) r=1
    (i)s +t=0
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by andy69 View Post
    so answering my questions
    (i) r=1
    (i)s +t=0
    Almost. r = -1, not 1. I hope you see how I matched the coefficients, because that's the main point of the problem, and you might get more problems like it.
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  7. #7
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    Quote Originally Posted by undefined View Post
    Almost. r = -1, not 1. I hope you see how I matched the coefficients, because that's the main point of the problem, and you might get more problems like it.
    yeah thanks i forgot the minus,great help anyway
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