Greatest possible value of a constant in polynomial

[songoku]

Homework Statement

A factor of $27x^18+bx^9+70$ is $px^9+q$. What is the greatest possible value of $b$?

Relevant Equations

Factor Theorem

Since $px^9+q$ is the factor, then $x^9=\frac{-q}{p}$ will be one of the roots.

Let $f(x)=27x^{18}+bx^9+70$, then:
$$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$
$$b=27 \frac{q}{p}+70 \frac{p}{q}$$
$$b=\frac{27q^2+70p^2}{pq}$$

From this expression, it looks like there is no greatest value of $b$ because increasing the value of $p$ and $q$ will also increase the value of $b$.

How to find the greatest value of $b$?

Thanks

 

[JimWhoKnew]

Are $b$, $p$, $q$ natural numbers?
Must the quotient polynomial have natural coefficients?
If so, you have a one parameter family of quadratic equations $~27y^2+Zy+70=0~$ . What are the values of $Z$ (real number) for which there is only one solution?
The answer suggests an upper bound for $b$ .

Edit: rephrased reply.

 

[Gavran]

Try to express b as a function f(y) where y=q/p and f(y)=27y+70(1/y). We can speak about the greatest value of b for q/p<0. If q/p>0, then we can speak about the smallest value of b.

 

[songoku]

Are $b$, $p$, $q$ natural numbers?
Must the quotient polynomial have natural coefficients?

I have posted the full question so I don't know about this

If so, you have a one parameter family of quadratic equations $~27y^2+Zy+70=0~$ . What are the values of $Z$ (real number) for which there is only one solution?
The answer suggests an upper bound for $b$ .

Sorry but why should we consider the case where the quadratic must have one solution?

Try to express b as a function f(y) where y=q/p and f(y)=27y+70(1/y). We can speak about the greatest value of b for q/p<0. If q/p>0, then we can speak about the smallest value of b.

Using differentiation, I can get the stationary point but I still don't see what the max value of $b$ since the greatest value of $b$ for $\frac{q}{p}<0$ is still less than min. value of $b$ for $\frac{q}{p}>0$

Thanks

 

[Gavran]

Using differentiation, I can get the stationary point but I still don't see what the max value of $b$ since the greatest value of $b$ for $\frac{q}{p}<0$ is still less than min. value of $b$ for $\frac{q}{p}>0$

You are right.
There are the local maximum and the local minimum. In my opinion, the question only holds for q/p<0.

 

[PeroK]

Homework Statement: A factor of $27x^18+bx^9+70$ is $px^9+q$. What is the greatest possible value of $b$?
Relevant Equations: Factor Theorem

Since $px^9+q$ is the factor, then $x^9=\frac{-q}{p}$ will be one of the roots.

Let $f(x)=27x^{18}+bx^9+70$, then:
$$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$
$$b=27 \frac{q}{p}+70 \frac{p}{q}$$
$$b=\frac{27q^2+70p^2}{pq}$$

From this expression, it looks like there is no greatest value of $b$ because increasing the value of $p$ and $q$ will also increase the value of $b$.

How to find the greatest value of $b$?

Thanks

This question makes no sense to me. If $p, q$ are fixed, then that determines $b$. There is no question of maximising $b$ in this case.

And, if we allow any $p, q$, then all that says is that we have at least one real root. That means that $b^2 > 4ac$ and there is no maximum for $b$. It can be as large as you like, and the polynomial will have real roots.

If $b > 0$, then there is a minimum of $b = 2\sqrt{ac}$, below which the polynomial has no real roots.

 

[FactChecker]

You are right. Given the problem as stated, there is no maximum value for $b$.

 

[songoku]

Thank you very much for the explanation JimWhoKnew, Gavran, PeroK, FactChecker

 

[fresh_42]

Homework Statement: A factor of $27x^18+bx^9+70$ is $px^9+q$. What is the greatest possible value of $b$?
Relevant Equations: Factor Theorem

Since $px^9+q$ is the factor, then $x^9=\frac{-q}{p}$ will be one of the roots.

Let $f(x)=27x^{18}+bx^9+70$, then:
$$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$
$$b=27 \frac{q}{p}+70 \frac{p}{q}$$
$$b=\frac{27q^2+70p^2}{pq}$$

From this expression, it looks like there is no greatest value of $b$ because increasing the value of $p$ and $q$ will also increase the value of $b$.

How to find the greatest value of $b$?

Thanks

$$
27x^{18}+bx^9+70=(px^9+q)\cdot \sum_{n=0}^9 a_nx^n=\sum_{n=9}^{18} pa_{n-9}x^{n}+\sum_{n=0}^9 qa_nx^n
$$
and therefore $70=qa_0\, , \,27=pa_9\, , \,b=pa_0+qa_9.$ There are only finitely many integer solutions for $(p,q,a_0,a_9)$ which can be tested easily.

Divisibility considerations only make sense over rings, not fields. Otherwise, we could as well complain about the fact that the polynomials weren't explicitly ruled out to be rational functions, which would allow any division.

 

[PeroK]

$$
27x^{18}+bx^9+70=(px^9+q)\cdot \sum_{n=0}^9 a_nx^n=\sum_{n=9}^{18} pa_{n-9}x^{n}+\sum_{n=0}^9 qa_nx^n
$$
and therefore $70=qa_0\, , \,27=pa_9\, , \,b=pa_0+qa_9.$ There are only finitely many integer solutions for $(p,q,a_0,a_9)$ which can be tested easily.

Divisibility considerations only make sense over rings, not fields. Otherwise, we could as well complain about the fact that the polynomials weren't explicitly ruled out to be rational functions, which would allow any division.

Alternatively, for integers $p, q, b$, wlog take $p = 1$, with $b = 27q + 70/q$.

For $b$ to be an integer, $q$ must divide 70. A quick look at the equation shows that $q = 70$ must give the largest value of $b$, namely 1891.