Use the Rational Zeros Theorem to find the possible zeros of the following polynomial function: #f(x)=33x^3-245x^2+407x-35#?

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Padria4mia · Accepted Answer

We have:            f              (        x        )              =    33          x      3        -    245          x      2        +    407    x    -    35  By the rational zeros theorem, any rational zeros of f(x) are expressible in the form             p      q       for integers p, q with p a divisor of the constant term -35 and q a divisor of the coefficient 33 of the leading term.Divisors of -35 include:      ±    1    ,    ±    5    ,    ±    7    ,    ±    35  The divisors of 33 are:      ±    1    ,    ±    3    ,    ±    11    ,    ±    33  These are some examples of rational zeros:      ±    1    ,    ±    5    ,    ±    7    ,    ±    35        ±          1      3        ,    ±          5      3        ,    ±          7      3        ,    ±          35      3            ±          1      11        ,    ±          5      11        ,    ±          7      11        ,    ±          35      11            ±          1      33        ,    ±          5      33        ,    ±          7      33        ,    ±          35      33      or in increasing order of size:      ±          1      33        ,    ±          1      11        ,    ±          5      33        ,    ±          7      33        ,    ±          5      11        ,    ±          7      11        ,    ±          1      3        ,    ±    1    ,    ±          35      33        ,    ±          5      3        ,    ±          7      3        ,    ±          35      11        ,    ±    5    ,    ±    7    ,    ±          35      3        ,    ±    35  Note that these are only the rational possibilities. The rational zeros theorem does not tell us about possible irrational or complex zeros.Using Descartes&#039; Rule of Signs, we can determine that this cubic has no negative zeros and 1 or 3 positive real zeros.So the only possible rational zeros are:            1      33        ,          1      11        ,          5      33        ,          7      33        ,          5      11        ,          7      11        ,          1      3        ,    1    ,          35      33        ,          5      3        ,          7      3        ,          35      11        ,    5    ,    7    ,          35      3        ,    35  Trying each in turn, we find:            f              (                  1          11                )              =    33                  (                                            1              11                                      )            3        -    245                  (                                            1              11                                      )            2        +    407          (                                    1            11                              )        -    35                                f                      (                          1              11                        )                                =                  3        -        245        +        4477        -        4235            121                                    f                      (                          1              11                        )                                =    0  So       x    =          1      11       is a zero and       11    x    -    1   a factor:      33          x      3        -    245          x      2        +    407    x    -    35    =          (      11      x      -      1      )              (      3              x        2            -      22      x      +      35      )      To factor the remaining quadratic we can use an AC method:Find a pair of factors of       A    C    =    3    ⋅    35    =    105   with sum       B    =    22  The pair 15, 7 works.Use this pair to split the middle term then factor by grouping:      3          x      2        -    22    x    +    35    =          (      3              x        2            -      15      x      )        -          (      7      x      -      35      )                          3                  x          2                -        22        x        +        35              =    3    x          (      x      -      5      )        -    7          (      x      -      5      )                          3                  x          2                -        22        x        +        35              =          (      3      x      -      7      )              (      x      -      5      )      So the other two zeros are:      x    =          7      3                                  and                                  x    =    5

Use the Rational Zeros Theorem to find the possible zeros of the following polynomial function: f(x)=33x^3-245x^2+407x-35?

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