Question

Let P(x) = x^(6)+ax^(5)+bx^(4)+cx^(3)+dx^(2)+ex+f.
If f is a prime number, how many distinct linear factors with integral coefficients can P(x) at most have?

Answer 1

@sirm3d would you kindly help me?

Answer 2

Try writing it as \[P(x) = (x - r_1)(x - r_2)(x - r_3)(x - r_4)(x - r_5)(x - r_6)\]

Then after expanding brackets you get
\[P(x) = x^6 + ... + r_1r_2r_3r_4r_5r_6\]

\[f = r_1r_2r_3r_4r_5r_6\]

Answer 3

wait...

Answer 4

common x ?

Answer 5

If r is a root to the function, then (x-r) is a linear factor
Since this is a sixth degree polynomial we have 6 roots, and we can write it as the product of the 6 linear factors

after multiplying out, you get that \[r_1r_2r_3r_4r_5r_6 = f\]
(ignoring all the other stuff)

if f is prime that means that there are at most 2 distinct linear factor with integral coefficients, which would be f and 1

Answer 6

ummm..

Answer 7

yup, and then

Answer 8

That's all since it answered the question :p

Answer 9

umm...

Answer 10

I still understand your point....

Answer 11

there will be at most three distinct linear factors, \[(x-1),(x+1),(x-f)\] or \[(x-1),(x+1),(x+f)\]

Answer 12

Oh derp I forgot the -1

Answer 13

actually, i don't understand the polynomials....

Answer 14

if f is prime, the only possible numbers to use in the synthetic division of p(x) by (x-r) are r=1, r=-1, r=f, or r=-f

Answer 15

if (x-f) or (x+f) is a factor of p(x), the constant in the quotient after the synthetic division is either 1 or -1. In this quotient, the possible factor is (x-1) or (x+1)

Answer 16

that means i should find out the quotient first, right?

Answer 17

the constant in the quotient is material in answering the question.

Answer 18

ok...