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

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)   (Show Source): You can put this solution on YOUR website!
.
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 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   ) 
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.

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