Rational Zeros of A Polynomial

Discussion in 'Other Pre-University Math' started by nycmathguy, Sep 20, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.5
    Question 96

    Can you work out 96 as a guide for me to do the rest? This is the final Section 2.5 question for us.

    20210919_213126.jpg
     
    nycmathguy, Sep 20, 2021
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  2. nycmathguy

    MathLover1

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    f(x)=x^3-(3/2)x^2-(23/2)x+6....factor out 1/2

    f(x)=(1/2)(2x^3-3x^2-23x+12)

    factor (2x^3-3x^2-23x+12)
    write
    -3x^2 as -x^2-2x^2
    and -23x as x-24x

    2x^3-x^2-2x^2+x-24x+12....group

    (2x^3-x^2)-(2x^2-x)-(24x-12)

    x^2(2x-1)-x(2x-1)-12(2x-1)

    (2x - 1) (x^2 - x - 12)

    now factor (x^2 - x - 12)
    x^2 - x - 12
    x^2 +3x- 4x - 12
    (x^2 +3x)- (4x +12)
    x(x +3)- 4(x +3)
    (x + 3) (x - 4)

    then you have
    f(x)=(2x - 1) (x + 3) (x - 4)
     
    MathLover1, Sep 20, 2021
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  3. nycmathguy

    nycmathguy

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    I can take it from here. Thank you.
     
    nycmathguy, Sep 20, 2021
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