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Math Help - PreCalcCaching Box 4 - Polynomials and Sigma help

  1. #1
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    Exclamation PreCalcCaching Box 4 - Polynomials and Sigma help

    I need the solutions to the following problems. I have worked hard on them and come up with answers but do not know how to proceed.
    1. Find the product of the roots of x^5-7x^2+6x-12=0
    My answer: 12.
    2. When x^27 - 6x^13 -x^4 + 4 is divided by x -1, the remainder is . What value will fill the empty box?
    My answer: -2
    3. In the equation 2x^4 - 6x^2 + 4x - 3 = 0, the sum of the roots is ___.
    My answer: 2 + sqrt(3)
    4. If (2 - i) and sqrt(3) are two roots of x^5 - 3x^4 - 2x^3 + 14x^2 - 3x - 15 = 0, find the equation's one rational root.
    My answer: -1.

    Final question to move on to next page: Evaluate
    I cannot determine an answer for this one.

    Thanks in advance
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  2. #2
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    I hate to be a pain but I need help on this ASAP... thanks
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    I do not understand what your question is. Are you having an issue with the last one? Could you be more clear?
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  4. #4
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    I agree with your answer to number 2. How did you get the rest? Are you willing to show your workings?
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  5. #5
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    Hello, khfan93!

    \text{1. Find the product of the roots of: }\;x^5-7x^2+6x-12\:=\:0

    \text{My answer: 12} . . . . Right!

    The leading coefficient must be 1 (one).
    If it is not 1, divide through by the leading coefficient.

    Write all the terms of the polynomial (insert 0 coefficients where needed).

    . . x^5 + 0x^4 + 0x^3 - 7x^2 + 6x - 12 \;=\;0

    Insert alternating signs on the coefficients (start with +).

    . . + \quad\; -  \quad\;\;\; + \quad\; - \quad\;\;\;+ \quad\;\;- <br />
    . . x^5 + 0x^4 + 0x^3 - 7x^2 + 6x - 12 \:=\:0
    . . . . . \Uparrow. . . . . . . . . . . . . . . . . \Uparrow
    . . . . sum of the roots. . . . . product of the roots


    Therefore, the product of the roots is 12.




    \text{2. When }x^{27} - 6x^{13} -x^4 + 4\text{ is divided by }(x -1),
    . . . \text{the remainder is: }\frac{\boxed{\,}}{x-1}

    \text{What value will fill the empty box?}

    \text{My answer: -2} . . . . Correct!

    When a polynomial P(x) is divided by (x-a), the remainder is: . P(a)


    Substitute x = 1 into the polynomial:

    . . P(1) \:=\:1^{27} - 6\cdot1^{13} - 1^4 + 4 \:=\:1 - 6 - 1 + 4 \;=\;-2




    \text{3. In the equation }2x^4 - 6x^2 + 4x - 3 \:=\: 0
    . . . \text{the sum of the roots is }\_\_.

    \text{My answer: }\:2 + \sqrt{3} . . . . no

    . . . . . . . . + \quad\;\; -
    We have: . x^4 + 0x^3 - 3x^2 + 2x - \frac{3}{2} \:=\:0
    . . . . . . . . . . . \Uparrow


    Therefore, the sum of the roots is 0.




    \text{4. If }(2 - i) \text{ and }\sqrt{3}\,\text{ are two roots of:}
    . . . . x^5 - 3x^4 - 2x^3 + 14x^2 - 3x - 15 \:=\: 0

    \text{find the equation's one rational root.}

    \text{My answer: -1} . . . . Yes!

    Complex roots always appear in conjugate pairs.
    If (2-i) is a root, then (2+i) is also a root.

    Irrational roots always appear in conjugate pairs.
    If \sqrt{3} is a root, then \text{-}\sqrt{3} is also a root.;

    So we have four of the roots; let the fifth root be r.


    In the equation, we see that the product of the roots is 15.

    So we have: . (2-i)(2+i)(\sqrt{3})(\text{-}\sqrt{3})r \;=\;15

    . . (5)(\text{-}3)r \:=\:15 \quad\Rightarrow\quad \text{-}15r \:=\:15 \quad\Rightarrow\quad r \:=\:\text{-}1

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