Divide Rational Expressions...2

Discussion in 'Algebra' started by nycmathguy, Aug 5, 2022.

  1. nycmathguy

    nycmathguy

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    College Algebra
    Chapter 1/Section 7

    IMG_20220805_185334.jpg

    IMG_20220805_185345.jpg

    IMG_20220805_185359.jpg
     
    nycmathguy, Aug 5, 2022
    #1
  2. nycmathguy

    HallsofIvy

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    Yes. Assuming you want to factor with integer coefficients then the factors of 9x^2+ 3x- 2 must have constant term 1, -2 or -1, 2. The coefficient of x must be 9, 1 or 3, 3 or -9,-1 or -3, -3.
    So possible factors are
    (9x+ 1)(x- 2)= 9x^2- 17x- 2
    (9x- 1)(x+ 2)= 9x^2+ 17xx- 2
    (x+ 1)(9x- 2)= 9x^2+ 7x- 2
    (x- 1)(9x+ 2)= 9x^2- 7x- 2
    (3x+ 1)(3x- 2)= 9x^2- 3x- 2
    (3x- 1)(3x+ 2)= 9x^2+ 3x- 2

    So the correct factorization is (3x- 1)(3x+ 2).
     
    HallsofIvy, Aug 16, 2022
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    nycmathguy likes this.
  3. nycmathguy

    nycmathguy

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    Looks good. Thanks again.
     
    nycmathguy, Aug 17, 2022
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