Finding m when x³ + mx² + 15x - 7 has rational roots

[uart]

Somebody gave me the following question, I was able to solve it but was unsure about some of the assumptions involved.

Question : Given that the polynomial x^3 + m x^2 + 15 x - 7 has at least two rational roots then find m.

Now the question didn't state that m had to be integer and I was unsure as to whether this was meant to be assumed or whether it could be deduced.

Here's what I did.

1. Since the product of the roots is 7 then two rational roots implies that the third root is also rational.

2. I assumed that m was integer which meant that the rational roots where also integer. ( by the http://planetmath.org/encyclopedia/RationalRootTheorem.html )

3. Since there are very few ways of having integer roots that multiply to give 7 I easily found the possible roots of 7, 1, 1, that multiply to give 7 and also have sum of pair-products totalling to 15.

4. So m = -(7 + 1 + 1) = -9


So did I need to assume that m was integer or could it have been deduced?

 

[matt grime]

You've found an m that satisfies the requirements. What does it matter that you assumed m was integer? It makes it easier and gives a unique solution, but so what? Had the assumption not yielded an answer then you couldn't conclude no solution existed, but that's all.

7=xyz
15=xy+xz+yz

15=xy+x(7/xy)+y(7/xy)=xy+7y+7x

so let x=2,

15=2y+7y+14, so y=1/9, and hence z=63/2 would also work, so it is neither necessary for m to be an integer, nor is it deducible that m is one.

 

[uart]

Yes I know that the solution is ok, and I'm sure it was the intended solution, but I am still uncertain as to whether it's unique.

Is there any theroem that would have implied that m had to be integer once I had all three roots rational, all other coefficients integer, and a monic polynomial ?

 

[matt grime]

See my example, in the above edited post. Clearly it is not necessary for m to be an integer, and cannot be deduced.

 

[uart]

Thanks Matt, that's just what I wanted to know. :)

 

[uart]

Hang on Matt, there is a mistake in that example.

It should be 15 = xy + 7/x + 7/y.

x=2 leads to a quadratic with irrational solutions for y.

 

[matt grime]

It's also clear that the choice of x=2 was completely arbitrary, and any rational would have yielded an answer.

 

[uart]

With the corrected equation (15 = xy + 7/x + 7/y), I don't think there is any possible rational solution for x and y other than the integer solution (x=1, y=1). I'm not sure how to prove that though.

 

[matt grime]

Damn, buggered that one up then.

need 15xy=x^2y^2 +7y+7x, so y is rational iff (7-15x)^2 - 28x^3 has rational square root (for some rational x) which seems too hard to solve, so I think we should assume that they meant that m was an integer, but let's try: we need to find a/b=x such that

(7b-15a)^2/b^2 - 28a^3/b^3 is a perfect square, If we let b be a square number, then it suffices (though is not necessary) that

b(7b-15a)^2-28a^3 is also a square. Anyone any suggestions?

In general though there is no theorem that states: if P is a monic poly with rational roots, then all but one coeff integer implies the last one must be: consider

x^2 - ax+1, If a=2, then there are two integer solutions, though in general

a=p+q where p=1/q and p is any rational will do, and I don't see it changing generally for higher degree polys: it really will depend on the poly in question.

 

[JasonRox]

What if m=-x? No one said it was a constant.

 

[JonF]

It was stated that it is a polynomial. And, polynomials are in the form of $$a_{n}x^{n} + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+...+a_{2}x^{2}+a_{1}x^{1}+a_{0}$$ where the a's are constant coefficients. So it is assumed that "m" in his problem would be a constant coefficient.