Barron's SAT Math 2 and SparkNotes state that if p/q is a rational zero of P(x) with integral coefficients, then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x) .
Then they show an example in which p/q isn't even a zero of the polynomial and continue to solve it as is (solving for 9/2 does not get you to 0):
$P(x) = 2x^4 + x^3 -19x^2 - 9x + 9$
Factors of constant term: $±1 , ±3 , ±9 $.
Factors of leading coefficient: $±1 , ±2 $.
Possible values of : $±1/1 , ±1/2 , ±3/1 , ±3/2 , ±9/1 , ±9/2 $. These can be simplified to: $±1 , ±1/2 , ±3 , ±3/2 , ±9 , ±9/2 $.
My questions are:
- Does p/q (where p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x)) need to be a zero of the polynomial in order to use this theorem? (since this obviously isn't the case in the above mentioned example and several others that I came across).
- What are the conditions for one to be able to use it?
- If anyone could provide an intuitive explanation/proof of why this theorem works in general that would be awesome.
1 Answer
$\begingroup$The theorem says if $p/q$ is a rational zero of $P(x)$, then .... The way you usually use this is to narrow down the possibilities for rational zeros. In this case $9/2$ does not happen to be a zero, but the theorem doesn't rule it out. Once you identify the possible rational zeros using the theorem, you can test each of them by substituting in to the polynomial and see which (if any) are actually zeros.
See e.g. Wikipedia for a proof.
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