The rational zeros theorem says that the roots of a polynomial P(x) will be of the form p/q where p is a factor of the constant coefficient and q is a factor of the leading coefficient. Why can't it be q/p? What's the reason that p/q is the right factor?
Rational Zeros Theorem
- Thread starter jpanknin
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Wow. You do not even have the theorem quite right.
Let’s start with some notation.
[math] P_n(x) \text { is an arbitrary polynomial of degree } n \ge 1 \text { with integer coefficients.} [/math]
Our proposition is that, if there are any rational zeroes, then one of those zeroes has p/q as a rational factor, where p is is an integer factor of the coefficient of the constant term and q is an integer factor is the coefficient of the leading term. The theorem DOES NOT SAY that there will be any rational zeroes. Nor does the theorem state that if there is a rational zero, that zero will necessarily equal the ratio p/q.
Now let’s take the case where n = 1. We have
[math]P_1(x) = qax + pb.\\ \text {If } P_1(x) = 0, \text { then } qax + pb = 0 \implies\\ - pb = qax = - pb \implies x = - \dfrac{pb}{qa} = \dfrac{p}{q} * \left ( - \dfrac{b}{a} \right ). [/math]
So we can see in this simple case very easily why the denominator is q: we need to divide by qa to solve for x.
Proving this in the general case is not so obvious, but it is the same general principle. At some point, we need to divide by the coefficient of [imath]x^n[/imath] and q is a factor of that coefficient. So q ends up in the denominator.
Does this at least begin to answer your question?
By the way, this simple case where n = 1 is special because if the coefficients are integers, the zero is necessarily rational. This is not true with higher order polynomials. In those more complex polynomials, the theorem applies only if there is a rational root.
Let’s start with some notation.
[math] P_n(x) \text { is an arbitrary polynomial of degree } n \ge 1 \text { with integer coefficients.} [/math]
Our proposition is that, if there are any rational zeroes, then one of those zeroes has p/q as a rational factor, where p is is an integer factor of the coefficient of the constant term and q is an integer factor is the coefficient of the leading term. The theorem DOES NOT SAY that there will be any rational zeroes. Nor does the theorem state that if there is a rational zero, that zero will necessarily equal the ratio p/q.
Now let’s take the case where n = 1. We have
[math]P_1(x) = qax + pb.\\ \text {If } P_1(x) = 0, \text { then } qax + pb = 0 \implies\\ - pb = qax = - pb \implies x = - \dfrac{pb}{qa} = \dfrac{p}{q} * \left ( - \dfrac{b}{a} \right ). [/math]
So we can see in this simple case very easily why the denominator is q: we need to divide by qa to solve for x.
Proving this in the general case is not so obvious, but it is the same general principle. At some point, we need to divide by the coefficient of [imath]x^n[/imath] and q is a factor of that coefficient. So q ends up in the denominator.
Does this at least begin to answer your question?
By the way, this simple case where n = 1 is special because if the coefficients are integers, the zero is necessarily rational. This is not true with higher order polynomials. In those more complex polynomials, the theorem applies only if there is a rational root.
Last edited:
Otis
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Hi jpanknin. The Rational Roots Theorem doesn't say that p/q is a factor of P(x).
It tells us: IF function P has a rational root, THEN that root will be of the form p/q as defined.
Have you worked any exercises using the Rational Roots Theorem, or have you seen a proof of it? To start investigating your question regarding form q/p, you could try using both forms p/q and q/p for comparison and see what happens.
[imath]\;[/imath]
Apologies for the delayed response. I've had covid and have been moving slowly.
Appreciate the replies and the clarification on the theorem, specifically the "IF there are any rational roots" part. I spent some time working through the proof of the theorem (pic included below from the book) and with the help of @Steven G's video and was able to isolate both [imath]p[/imath] and [imath]q[/imath], showing that [imath]p[/imath] is a factor of [imath]a_0[/imath] and [imath]q[/imath] is a factor of [imath]a_n[/imath].
I get the right answer when I try all the [imath]\frac{p}{q}[/imath] combinations, so I know it works, but I can't see WHY [imath]\frac{p}{q}[/imath] is right and [imath]\frac{q}{p}[/imath] is not. I've spent a lot of time watching videos and reading articles and most work out isolating [imath]p[/imath] and [imath]q[/imath], meaning showing that [imath]\frac{a_0q^n}{p}[/imath] and [imath]\frac{a_np^n}{q}[/imath], but stop short of explaining why every rational zero of a polynomial will be of the form where [imath]p[/imath] is the numerator and [imath]q[/imath] the denominator.
@JeffM , I feel like this line from your reply is key, but I'm still not seeing it: "Proving this in the general case is not so obvious, but it is the same general principle. At some point, we need to divide by the coefficient of [imath]x^n[/imath] and q is a factor of that coefficient. So q ends up in the denominator."
Appreciate the replies and the clarification on the theorem, specifically the "IF there are any rational roots" part. I spent some time working through the proof of the theorem (pic included below from the book) and with the help of @Steven G's video and was able to isolate both [imath]p[/imath] and [imath]q[/imath], showing that [imath]p[/imath] is a factor of [imath]a_0[/imath] and [imath]q[/imath] is a factor of [imath]a_n[/imath].
I get the right answer when I try all the [imath]\frac{p}{q}[/imath] combinations, so I know it works, but I can't see WHY [imath]\frac{p}{q}[/imath] is right and [imath]\frac{q}{p}[/imath] is not. I've spent a lot of time watching videos and reading articles and most work out isolating [imath]p[/imath] and [imath]q[/imath], meaning showing that [imath]\frac{a_0q^n}{p}[/imath] and [imath]\frac{a_np^n}{q}[/imath], but stop short of explaining why every rational zero of a polynomial will be of the form where [imath]p[/imath] is the numerator and [imath]q[/imath] the denominator.
@JeffM , I feel like this line from your reply is key, but I'm still not seeing it: "Proving this in the general case is not so obvious, but it is the same general principle. At some point, we need to divide by the coefficient of [imath]x^n[/imath] and q is a factor of that coefficient. So q ends up in the denominator."
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I think I may have gotten it. Please let me know if I'm on the right track.
If we have the polynomial [imath]P(x) = 2x^3+7x^2+4x-4[/imath], we can factor it into [imath](2x-1)(x+2)^2[/imath], so the zeros will be [imath]\frac{1}{2}[/imath] and [imath]-2[/imath] with a multiplicity of 2. The [imath]\frac{1}{2}[/imath] was obtained by dividing by 2, which is the coefficient and a factor of [imath]a_n[/imath]. So when the coefficient of [imath]a_n[/imath] is not 1, we'll eventually have to divide by that coefficient (or a factor of that coefficient) to solve for [imath]x[/imath] (i.e., [imath]2x-1=0[/imath]). Therefore, [imath]q[/imath], as a factor of [imath]a_n[/imath], will necessarily be the denominator in any rational zero, if it exists. Right track?
If we have the polynomial [imath]P(x) = 2x^3+7x^2+4x-4[/imath], we can factor it into [imath](2x-1)(x+2)^2[/imath], so the zeros will be [imath]\frac{1}{2}[/imath] and [imath]-2[/imath] with a multiplicity of 2. The [imath]\frac{1}{2}[/imath] was obtained by dividing by 2, which is the coefficient and a factor of [imath]a_n[/imath]. So when the coefficient of [imath]a_n[/imath] is not 1, we'll eventually have to divide by that coefficient (or a factor of that coefficient) to solve for [imath]x[/imath] (i.e., [imath]2x-1=0[/imath]). Therefore, [imath]q[/imath], as a factor of [imath]a_n[/imath], will necessarily be the denominator in any rational zero, if it exists. Right track?
Yes, you are on the right track.
I shall try to give a more complete response later, but it is Christmas so of course my daughter, my one-year old grandson, and my wife all have colds. There may be peace on earth generally, but not in my house.
Otis
Elite Member
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There's a good catch at timestamp 4-seconds.
Nice video, by the way.
[imath]\;[/imath]