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Finding a 4th degree polynomial

  • #1
Problem:

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

I found that the two roots of [itex]q(x)[/itex] are [itex]x=7\sqrt{2}-\sqrt{11}[/itex] and [itex]x=7\sqrt{2}+\sqrt{11}[/itex]. 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 [itex]14\sqrt{2}[/itex] like so:

[tex](x^2-14\sqrt{2}+87)(x+14\sqrt{2})[/tex] but I ended up with
[tex](x^3+87x+1218\sqrt{2}-392)[/tex] 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?
 

Answers and Replies

  • #2
Ray Vickson
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Problem:

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

I found that the two roots of [itex]q(x)[/itex] are [itex]x=7\sqrt{2}-\sqrt{11}[/itex] and [itex]x=7\sqrt{2}+\sqrt{11}[/itex]. 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 [itex]14\sqrt{2}[/itex] like so:

[tex](x^2-14\sqrt{2}+87)(x+14\sqrt{2})[/tex] but I ended up with
[tex](x^3+87x+1218\sqrt{2}-392)[/tex] 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
[tex] p(x) = (a x^2 + bx + c) q(x).[/tex]
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.
 
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  • #3
Ok, now I have expanded it out as you suggested:
[tex]p(x)=(ax^2+bx+c)(x^2-14\sqrt{2}x+87)[/tex]
which ends up with:
[tex]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[/tex]

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...)
 
  • #4
Ray Vickson
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Ok, now I have expanded it out as you suggested:
[tex]p(x)=(ax^2+bx+c)(x^2-14\sqrt{2}x+87)[/tex]
which ends up with:
[tex]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[/tex]

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
[tex] p(x) = N_4 x^4 + N_3 x^3 + N_2 x^2 + N_1 x + N_0,[/tex]
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.)
 
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