Finding a 4th degree polynomial

[SwimmingGoat]

Problem:

q(x)=x^2-14\sqrt{2}x+87. Find 4th degree polynomial p(x) with integer coefficients whose roots include the roots of q(x). What are the other two roots of p(x)?

I found that the two roots of q(x) are x=7\sqrt{2}-\sqrt{11} and x=7\sqrt{2}+\sqrt{11}. Since they are conjugates of each other, I have no idea what to guess the other roots could be of my fourth-degree polynomial.

I started out with trying to get rid of the 14\sqrt{2} like so:

(x^2-14\sqrt{2}+87)(x+14\sqrt{2}) but I ended up with
(x^3+87x+1218\sqrt{2}-392) Going to the fourth degree looked like a headache, and I felt I wasn't on the right track, so I stopped there.

Any ideas?

 

[Ray Vickson]

Problem:

q(x)=x^2-14\sqrt{2}x+87. Find 4th degree polynomial p(x) with integer coefficients whose roots include the roots of q(x). What are the other two roots of p(x)?

I found that the two roots of q(x) are x=7\sqrt{2}-\sqrt{11} and x=7\sqrt{2}+\sqrt{11}. Since they are conjugates of each other, I have no idea what to guess the other roots could be of my fourth-degree polynomial.

I started out with trying to get rid of the 14\sqrt{2} like so:

(x^2-14\sqrt{2}+87)(x+14\sqrt{2}) but I ended up with
(x^3+87x+1218\sqrt{2}-392) Going to the fourth degree looked like a headache, and I felt I wasn't on the right track, so I stopped there.

Any ideas?

Since the roots of p(x) include those of q(x), p(x) must be a multiple of q(x); that is, we must have
p(x) = (a x^2 + bx + c) q(x).
Expand out p(x) and equate the coefficients of the $x^j$ to integers; this will give some restrictions on a, b and c, and you can start your search from there.

 

[SwimmingGoat]

Ok, now I have expanded it out as you suggested:
p(x)=(ax^2+bx+c)(x^2-14\sqrt{2}x+87)
which ends up with:
p(x)=(a)x^4+(b-14\sqrt{2}a)x^3+(87a-14\sqrt{2}b+c)x^2+(87b-14\sqrt{2}c)x+87c

From here, do I try to make educated guesses for a,b, and c? Or do these restrictions give some obvious clues? (I'm sorry if the questions seem easy, I haven't done math in a couple of years, and I'm quite rusty...)

 

[Ray Vickson]

Ok, now I have expanded it out as you suggested:
p(x)=(ax^2+bx+c)(x^2-14\sqrt{2}x+87)
which ends up with:
p(x)=(a)x^4+(b-14\sqrt{2}a)x^3+(87a-14\sqrt{2}b+c)x^2+(87b-14\sqrt{2}c)x+87c

From here, do I try to make educated guesses for a,b, and c? Or do these restrictions give some obvious clues? (I'm sorry if the questions seem easy, I haven't done math in a couple of years, and I'm quite rusty...)

If you write
p(x) = N_4 x^4 + N_3 x^3 + N_2 x^2 + N_1 x + N_0,
where the $N_i$ are integers, you can solve for a,b,c in terms of $N_4,N_3,N_2$. Then you can use those formulas to express $N_0$ and $N_1$ in terms of $N_4,N_3,N_2$. By requiring that $N_0$ and $N_1$ be integers, this will greatly restrict the possible values of $N_2,N_3,N_4$. Try it and see! (Admittedly, I used the computer algebra package Maple to do all the work, but I guess you might be able to do it using Wolfram Alpha---available freely on-line. Even doing it by hand is not too bad.)