Factor Theorem...4

Discussion in 'Other Pre-University Math' started by nycmathguy, Sep 11, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.3
    Question 60

    x^3 + 2x^2 - 2x - 4 = 0 given x = sqrt{2}.

    If x = sqrt{2} leads to 0, one factor of the polynomial is x - sqrt{2}.

    Using synthetic division, I came up with
    x^2 + (2 + 2sqrt{2}) + 2sqrt{2}.

    Factoring I got (x + 2)(x + sqrt{2}).

    List zeros:

    x = -2, x = -sqrt{2}

    You say?
     
    nycmathguy, Sep 11, 2021
    #1
  2. nycmathguy

    MathLover1

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    you forgot x
    x^2 + (2 + sqrt(2))x + 2sqrt(2).........factor
    x^2 + 2x + 2sqrt(2)x + 2sqrt(2)
    (x^2 + 2sqrt(2)x )+ (2x + 2sqrt(2))
    x(x + 2sqrt(2) )+ 2(x + sqrt(2))
    (x+ 2)(x + sqrt(2))

    => x^3 + 2x^2 - 2x - 4 = (x - sqrt(2))(x+ 2)(x + sqrt(2))

    List zeros:
    x =sqrt(2)
    x=- 2
    x =-sqrt(2)
     
    MathLover1, Sep 11, 2021
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Not only did I forget x, I also forgot x = sqrt{2} as a third zero of the polynomial. I may have to read Ron Larson's chapter notes again. I started Section 2.3 in the middle of the moving process from Queens to Brooklyn.

    This is not an excuse. I have been very stressed out lately. I just found out that a very painful procedure called prostate biopsy is needed before the year ends. This means that there's a good chance I may have prostate cancer. Prostate cancer is > math.
     
    nycmathguy, Sep 11, 2021
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