Remainder Theorem...3

Discussion in 'Other Pre-University Math' started by nycmathguy, Sep 9, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.3
    Question 48

    20210903_125953.jpg

    If zero is at k= 1/5, one factor of the given polynomial is (x-(1/5))=(x - 1/5).

    (1/5)(10) = 2

    -22 + 2 = -20

    (1/5)(-20) = -4

    -3 + (-4) = -7

    (1/5)(-7) = -7/5

    -7/5 + 4 = 13/5

    I say the answer is f(x) = 2x^2 - 20x - 7 with
    remainder 13/5.

    You say?
     
    nycmathguy, Sep 9, 2021
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  2. nycmathguy

    MathLover1

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    48.
    10x^3-22x^2-3x +4=0
    x=1/5

    ..1/5.| 10.................. -22..................-3........................ 4

    ......... ...... (1/5)10=2.......(1/5)(-20)=-4.....(1/5)(-7)=-7/5
    _________________________________________
    .........10..................-20..................-7.........................13/5

    resulting coefficients: 10,-20,-7, 13/5

    the quotient is 10x^2-20x-7, and the remainder is 13/5
     
    MathLover1, Sep 9, 2021
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    I did not include 10 in the list of coefficients. I will try again. For the most part, not too bad.
     
    nycmathguy, Sep 9, 2021
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    MathLover1 likes this.
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