Factorization of a polynomial into linear and quadratic factors












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It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?






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    It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?






    polynomials factoring alternative-proof







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      It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?






      polynomials factoring alternative-proof







      share|cite|improve this question















      It is known that every non-constant polynomial with real coefficients admits a factorization in terms of real and quadratic factors. The proof normally uses the Fundamental Theorem of Algebra. Is there an elementary proof of the above which does not involve complex numbers at all?






      polynomials factoring alternative-proof




      polynomials factoring alternative-proof






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      edited Dec 3 '18 at 14:20









      José Carlos Santos

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      asked Feb 16 '18 at 17:56









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          I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.



          So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.



          In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.



          It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:




          1. $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.

          2. $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.






          share|cite|improve this answer























          • Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
            – user477343
            Feb 18 '18 at 12:24













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          I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.



          So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.



          In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.



          It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:




          1. $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.

          2. $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.






          share|cite|improve this answer























          • Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
            – user477343
            Feb 18 '18 at 12:24


















          1














          I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.



          So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.



          In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.



          It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:




          1. $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.

          2. $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.






          share|cite|improve this answer























          • Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
            – user477343
            Feb 18 '18 at 12:24
















          1












          1








          1






          I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.



          So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.



          In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.



          It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:




          1. $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.

          2. $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.






          share|cite|improve this answer














          I published such a proof (see my article Another Proof of the Fundamental Theorem of Algebra, American Mathematical Monthly 112 (1), 2005, pp. 76–78), although I doubt that you'll find it elementary.



          So, let $p(x)inmathbb{R}[x]$ be an irreducible polynomial. If I prove that its degree is $1$ or $2$, I will have proved that every polynomial in $mathbb{R}[x]$ can be written as a product of linear and quadratic polynomials. Let $n=deg p(x)$ and assume that $n>1$. The idea is to prove that $n=2$. Note that $mathbb{R}[x]/bigllangle p(x)bigrrangle$ is a field which is an extension of $mathbb R$ and whose dimension as a $mathbb R$-vector space is $n$. So, all that is needed is to prove that if $K$ is such an extension of $mathbb R$, then $n=2$.



          In order to prove that, I proved that there is a norm $|cdot|$ on $K$ such that$$(forall x,yin K):|x.y|leqslant|x|.|y|.$$This allows us to define the exponential function$$begin{array}{rccc}expcolon&K&longrightarrow&K\&x&mapsto&displaystylesum_{n=0}^inftyfrac{x^n}{n!}.end{array}$$It turns out that $(forall x,yin K):exp(x+y)=exp(x)exp(y)$. That is, $exp$ is a group homomorphism from $(K,+)$ into $(Ksetminus{0},.)$. It can be proved that $exp(K)$ is both an open and a closed set of $Ksetminus{0}$. Since $n>1$, $Ksetminus{0}$ is connected and therefore $exp$ is surjective.



          It can also be proved that $kerexp$ is a discrete subgroup of $(K,+)$. This means that either $kerexp={0}$ or that there are $k$ linearly independent vectors $v_1,ldots,v_kin K$ such that $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. It is not hard to prove that $exp$ induces a homeomorphism between $K/kerexp$ and $Ksetminus{0}$. But then there are two possibilites:




          1. $kerexp={0}$: this is impossible, because $mathbb{R}^n$ and $mathbb{R}^nsetminus{0}$ are not homeomorphic.

          2. $kerexp=mathbb{Z}v_1opluscdotsoplusmathbb{Z}v_k$. Then $K/kerexp$ is homeomorphic to $(S^1)^ktimesmathbb{R}^{n-k}$, which is not simply connected. But $mathbb{R}^nsetminus{0}$ is simply connected when $n>2$. Therefore, $n=2$ and this completes my (not that much elementary) proof.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 19 '18 at 10:33

























          answered Feb 18 '18 at 12:21









          José Carlos Santos

          150k22122223




          150k22122223












          • Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
            – user477343
            Feb 18 '18 at 12:24




















          • Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
            – user477343
            Feb 18 '18 at 12:24


















          Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
          – user477343
          Feb 18 '18 at 12:24






          Well done for actually publishing a proof. I want to publish something in the following years, whether it be a proof, theorem, conjecture, or improvement on a certain discovery (like reducing a well-known bound or something), etc.
          – user477343
          Feb 18 '18 at 12:24




















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