True/false: $f(x)=x^4-x^3+14x^2+5x+16$ is the product of two degree two polynomials over $Bbb Z$?












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






polynomials







share|cite|improve this question




















  • 3




    you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$
    – Will Jagy
    Nov 30 at 0:13










  • @WillJagy give me 3 minutes im calculating
    – jasmine
    Nov 30 at 0:14






  • 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 at 0:23








  • 1




    Try searching Eisenstein's theorem.
    – Oolong milk tea
    Nov 30 at 0:26






  • 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 at 0:30
















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






polynomials







share|cite|improve this question




















  • 3




    you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$
    – Will Jagy
    Nov 30 at 0:13










  • @WillJagy give me 3 minutes im calculating
    – jasmine
    Nov 30 at 0:14






  • 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 at 0:23








  • 1




    Try searching Eisenstein's theorem.
    – Oolong milk tea
    Nov 30 at 0:26






  • 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 at 0:30














2












2








2


1






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






polynomials







share|cite|improve this question
















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






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 0:13









Shaun

8,329113578




8,329113578










asked Nov 30 at 0:07









jasmine

1,519416




1,519416








  • 3




    you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$
    – Will Jagy
    Nov 30 at 0:13










  • @WillJagy give me 3 minutes im calculating
    – jasmine
    Nov 30 at 0:14






  • 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 at 0:23








  • 1




    Try searching Eisenstein's theorem.
    – Oolong milk tea
    Nov 30 at 0:26






  • 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 at 0:30














  • 3




    you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$
    – Will Jagy
    Nov 30 at 0:13










  • @WillJagy give me 3 minutes im calculating
    – jasmine
    Nov 30 at 0:14






  • 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 at 0:23








  • 1




    Try searching Eisenstein's theorem.
    – Oolong milk tea
    Nov 30 at 0:26






  • 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 at 0:30








3




3




you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$
– Will Jagy
Nov 30 at 0:13




you have begun properly. Using your letters $a,b,c,d,$ what is the required value for $a+c?$
– Will Jagy
Nov 30 at 0:13












@WillJagy give me 3 minutes im calculating
– jasmine
Nov 30 at 0:14




@WillJagy give me 3 minutes im calculating
– jasmine
Nov 30 at 0:14




1




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 at 0:23






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 at 0:23






1




1




Try searching Eisenstein's theorem.
– Oolong milk tea
Nov 30 at 0:26




Try searching Eisenstein's theorem.
– Oolong milk tea
Nov 30 at 0:26




1




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 at 0:30




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 at 0:30










1 Answer
1






active

oldest

votes


















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 $mathbb{Z}[x]$?






share|cite|improve this answer





















  • why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
    – jasmine
    Nov 30 at 0:23








  • 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 at 0:29










  • okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
    – jasmine
    Nov 30 at 0:31






  • 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 at 0:40










  • @DanielSchepler thanks u, got its
    – jasmine
    Nov 30 at 0:44











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1 Answer
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active

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1 Answer
1






active

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active

oldest

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active

oldest

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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 $mathbb{Z}[x]$?






share|cite|improve this answer





















  • why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
    – jasmine
    Nov 30 at 0:23








  • 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 at 0:29










  • okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
    – jasmine
    Nov 30 at 0:31






  • 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 at 0:40










  • @DanielSchepler thanks u, got its
    – jasmine
    Nov 30 at 0:44
















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 $mathbb{Z}[x]$?






share|cite|improve this answer





















  • why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
    – jasmine
    Nov 30 at 0:23








  • 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 at 0:29










  • okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
    – jasmine
    Nov 30 at 0:31






  • 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 at 0:40










  • @DanielSchepler thanks u, got its
    – jasmine
    Nov 30 at 0:44














6












6








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 $mathbb{Z}[x]$?






share|cite|improve this answer












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 $mathbb{Z}[x]$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 30 at 0:22









Daniel Schepler

8,2641618




8,2641618












  • why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
    – jasmine
    Nov 30 at 0:23








  • 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 at 0:29










  • okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
    – jasmine
    Nov 30 at 0:31






  • 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 at 0:40










  • @DanielSchepler thanks u, got its
    – jasmine
    Nov 30 at 0:44


















  • why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
    – jasmine
    Nov 30 at 0:23








  • 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 at 0:29










  • okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
    – jasmine
    Nov 30 at 0:31






  • 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 at 0:40










  • @DanielSchepler thanks u, got its
    – jasmine
    Nov 30 at 0:44
















why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
– jasmine
Nov 30 at 0:23






why u take $mathbb{Z_2}$?? question is about $mathbb{Z}$?
– jasmine
Nov 30 at 0:23






1




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 at 0:29




@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 at 0:29












okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
– jasmine
Nov 30 at 0:31




okkss @CarlSchildkraut..thanks u i gots its .... can u said me that which theorem is this ?
– jasmine
Nov 30 at 0:31




1




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 at 0:40




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 at 0:40












@DanielSchepler thanks u, got its
– jasmine
Nov 30 at 0:44




@DanielSchepler thanks u, got its
– jasmine
Nov 30 at 0:44


















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