2

Is the following statement is true/false?

Consider the polynomial $$f(x)=x^4-x^3+14x^2+5x+16,$$ then $f$ is a product of two polynomials of degree two over $\mathbb{Z}$.

My answer : I think it will be true $f(x)=x^4-x^3+14x^2+5x+16= (x^2 +ax+b)(x^2 +cx + d)$

Is its True ?

Any hints/solution will be appreciated

Community
  • 1
jasmine
  • 14,457
  • 3
    you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$ – Will Jagy Nov 30 '18 at 00:13
  • @WillJagy give me 3 minutes im calculating – jasmine Nov 30 '18 at 00:14
  • I actually can wait... however, what I do is to multiply out JUST ENOUGH to figure out $a+c.$ Then I go on to another easy one, such as what is the product $bd ; ? ; ;$ My point is, mutlitplying out the whole thing in symbols makes a bit of a mess; – Will Jagy Nov 30 '18 at 00:17
  • @WillJagy $a+ c = -1$ – jasmine Nov 30 '18 at 00:18
  • I thought of a good one, since all must be integers. First, what is $bd ; ? ;$ Second, what is $ad+bc ?$ The two facts put together make something nice using odd/even numbers. – Will Jagy Nov 30 '18 at 00:21
  • bd =16 @ Will jagy – jasmine Nov 30 '18 at 00:22
  • 1
    Alright, a caution: either both $b,d > 0$ or both $b,d < 0.$ Anyway, $bd$ is even. Now, what is $ad+bc ; ? ;$ And, this is the first punchline, can $b,d$ both be even? – Will Jagy Nov 30 '18 at 00:23
  • 1
    Try searching Eisenstein's theorem. – Oolong Milktea Nov 30 '18 at 00:26
  • 1
    too many people; as far as we had gotten, as $ad+bc$ is odd, we find that one of $b,d$ is odd, so the total possibilities for ordered $(b,d)$ are $(1,16),$ $(-1,-16),$ $(16,1),$ $(-16,-1),$ For each pair one may, well, continue with $c=(1-a)$ written in, see what happens – Will Jagy Nov 30 '18 at 00:30
  • @WillJagy sir can u explain ur answer in detail in answer BOX – jasmine Nov 30 '18 at 00:35

1 Answers1

6

Hint: the image of $f(x)$ in $(\mathbb{Z} / 2 \mathbb{Z})[x]$ is $x^4 + x^3 + x = x(x^3 + x^2 + 1)$. However, $x^3 + x^2 + 1$ is irreducible in this ring (why?). Can you complete the proof to show that this implies that the original polynomial cannot be a product of two quadratic polynomials in $\mathbb{Z}[x]$?

Daniel Schepler
  • 20,364
  • why u take $\mathbb{Z_2}$?? question is about $\mathbb{Z}$? – jasmine Nov 30 '18 at 00:23
  • 1
    @jasmine Show that if a polynomial were a product of two quadratics in $\mathbb{Z}[x]$, it would also be in $\mathbb{Z}_2[x]$. – Carl Schildkraut Nov 30 '18 at 00:29
  • okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ? – jasmine Nov 30 '18 at 00:31
  • 1
    I don't think there's any particular theorem - just the fact that the function $\mathbb{Z}[x] \to (\mathbb{Z} / 2 \mathbb{Z})[x]$, $\sum a_i x^i \mapsto \sum \overline{a_i} x^i$, is a ring homomorphism. (Do note, however, that the proof will make use in an essential way of the fact that $(\mathbb{Z} / 2 \mathbb{Z})[x]$ is a unique factorization domain - for which you probably do have a theorem to cite to justify this fact.) – Daniel Schepler Nov 30 '18 at 00:40
  • @DanielSchepler thanks u, got its – jasmine Nov 30 '18 at 00:44