## Polynomial problems

Forum for the GRE subject test in mathematics.
lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am

### Polynomial problems

1. For what integer "a" x^13+x+90 is divisible by x^2-x+a ?

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am

### Simplest technique

If f(x) = x^13 + x + 90 and p(x) = x^2 - x + a, we want to find a so that there exists a q(x) such that f = p * q.

The simplest way to find this is set x to 0, -1, and 1 and deduce the value of a. Set x = 0 and a|90, set x = 1 and a|92, and set x= -1 and 2+a|88. The GCD of 90 and 92 is 2, and if a = 1, 3 would not divide 88. So a = 2.

Otherwise, the best technique (which I'm quite certain is not allowed on the GRE) is to divide the polynomials in Mathematica until a reducible polynomial is found. lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
Set x = 0 and a|90
How come you deduce from it that a must divide 90?

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am
My reasoning is that x^13 +x +90 is an element of Z[x] which is the polynomial ring with integer coefficients, so it must be closed under multiplication. The Euclidean algorithm would imply that a | 90.

Also, there's only one constant term in the multiplication of p(x)*q(x) which would be a*k = 90, so for an integer a, k too must be an integer and a | 90.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
My reasoning is that x^13 +x +90 is an element of Z[x]
Yeah. That's exactly what I was afraid of. The problem doesn't say that it is Z[x], not Q[x]. But, apparently, you're right, it must be Z[x], otherwise I cannot see how to solve it. Thanks for solution!

Here is one more.

Find all polynomials P(x) which satisfy:
x*P(x-1)=(x-26)*P(x)

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am
Find all polynomials P(x) which satisfy:
x*P(x-1)=(x-26)*P(x)

I took a similar approach to this as I did in the last problem, and there are multiple ways to solve this.

Let x = 0; 0*P(-1) = -26*P(0) iff P(0) = 0, so this implies that the constant coefficient of any polynomial that satisfies this equation will be 0.

So, the form of P(x) must be a*x^n + ... z*x + 0 which implies that x | P(x), so (x-26) must divide P(x-1) for P(x-1)/(x-26) = P(x)/x.

Then if we let x = 1, P(0)/(-25) = P(1) ==> P(1) = 0
x = 2, P(1)/(-24) = P(2)/2 ==> P(2) = 0, and so forth, so...

The zero polynomial must be the only polynomial to satisfy that relation.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
so (x-26) must divide P(x-1)
Ok.
Then if we let x = 1, P(0)/(-25) = P(1) ==> P(1) = 0
x = 2, P(1)/(-24) = P(2)/2 ==> P(2) = 0, and so forth, so...
Continuing, we have
P(24)=0;
P(25)=0.
...
Then if we let x=26, we would have

26*P(25)=0*P(26) or 0=0.
And we cannot conclude from it that P(26)=0 or can we?

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am
Abel's theorem states that a quintic polynomial cannot be solved entirely in radicals, or real numbers (1 is the same as writing 1 + sqrt (0)) .

The statement of the theorem from Wolfram's website:

In general, polynomial equations higher than fourth degree are incapable of algebraic solution in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions. This was also shown by Ruffini in 1813 (Wells 1986, p. 59).

So, as soon as we can show that P(0) = P(1) = ... = P(4) = 0, we know that the polynomial must be the zero polynomial, since a polynomial of any (positive) degree can have at most 4 roots solved in radicals. Apparently, the degree of the zero polynomial is set to -inf.

There was one problem on test 0568 that relates to this theorem #17.

How many real roots does the polynomial 2x^5 +8x - 7 have?

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
as soon as we can show that P(0) = P(1) = ... = P(4) = 0, we know that the polynomial must be the zero polynomial
You are not serious, are you?
Counterexample:
P(x) = x(x-1)(x-2)(x-3)(x-4) - not zero polynomial
How many real roots does the polynomial 2x^5 +8x - 7 have?
It can be solved by figuring out that derivative is non negative everywhere.

zombie
Posts: 27
Joined: Thu Nov 20, 2008 2:30 am

### oops!

Yeah, I forgot constructions like (x+1)^100 which has 100 repeated roots of -1.

I misinterpreted Abel's theorem. It only applies to the polynomial ring or field operations, not to synthetic division of the polynomial.

I had been confounding the conjugate roots theorems with Abel's theorem! Thanks for the simple counterexample.

However, I think I'm still right about the zero polynomial being the only P(x) to satisfy that relation.