## Polynomial problems

### Polynomial problems

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

### 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

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.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.

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!My reasoning is that x^13 +x +90 is an element of Z[x]

Here is one more.

**Find all polynomials P(x) which satisfy:**

x*P(x-1)=(x-26)*P(x)

x*P(x-1)=(x-26)*P(x)

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.

Ok.so (x-26) must divide P(x-1)

Continuing, we haveThen 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...

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?

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?**

You are not serious, are you?as soon as we can show that P(0) = P(1) = ... = P(4) = 0, we know that the polynomial must be the zero polynomial

Counterexample:

P(x) = x(x-1)(x-2)(x-3)(x-4) - not zero polynomial

It can be solved by figuring out that derivative is non negative everywhere.How many real roots does the polynomial 2x^5 +8x - 7 have?

### 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.

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.