SOLUTION: If the polynomial function P(x) = an^x^n + an−1^x^n−1 + ... + a1^x + a0 has integer coefficients, then the only numbers that could possibly be rational zeros of P are all o

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: If the polynomial function P(x) = an^x^n + an−1^x^n−1 + ... + a1^x + a0 has integer coefficients, then the only numbers that could possibly be rational zeros of P are all o      Log On


   



Question 1200950: If the polynomial function
P(x) = an^x^n + an−1^x^n−1 + ... + a1^x + a0
has integer coefficients, then the only numbers that could possibly be rational
zeros of P are all of the form p/q, where p is a factor of the constant coefficient a0 and q is a factor of the leading coefficient an

The possible rational zeros of P(x) = 10x^3 + 6x^2 − 21x − 34 are:

Answer by ikleyn(52781) About Me  (Show Source):
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If the polynomial function P(x) = an^x^n + an−1^x^n−1 + ... + a1^x + a0
has integer coefficients, then the only numbers that could possibly be rational zeros of P
are all of the form p/q, where p is a factor of the constant coefficient a0
and q is a factor of the leading coefficient an
The possible rational zeros of P(x) = 10x^3 + 6x^2 − 21x − 34 are:
~~~~~~~~~~~~~~~~~~~~~


            In this problem,  they do not ask you to find the zeros.
            They want you form the list of highlight%28possible%29 rational zeroes,  based on the described algorithm.


So, based on the described algorithm, you write the list of all integer factors of 
the constant term p= a0 = -34.  These factors are

    +/- 1, +/- 2, +/- 17, +/- 34.


Next, you write the list of all factors of the leading coefficients q= an = 10.
These factors are

    +/- 1, +/- 2, +/- 5, +/- 10.


Now the list of all possible rational zeroes ( the list of all  highlight%28CANDIDATES%29 ) 
is the long set of all possible ratios

   { +/- 1,    +/- 2,    +/- 17,   +/- 34,

     +/- 1/2,  +/- 2/2,  +/-17/2,  +/- 34/2,

     +/- 1/5,  +/- 2/5,  +/-17/5,  +/- 34/5,

     +/- 1/10, +/- 2/10, +/-17/10, +/- 34/10 }.


The last step is to reduce / (to simplify) the fractions - where it is possible, 
and remove the repeating terms, that can arise after reducing. 
So, the final list is this

   { +/- 1,    +/- 2,    +/- 17,    +/- 34,

     +/- 1/2,            +/- 17/2,  

     +/- 1/5,  +/- 2/5,  +/- 17/5,  +/- 34/5,

     +/- 1/10,           +/- 17/10           }.

Solved.