What's the intuition behind the fact that you can factorise an expression as $(x-a)(x-b)$ if $a$ and $b$ are the roots of the expression

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Say $f(x)$ is a polynomial function with roots $a, , b, , c$ then this can be expressed as $f(x) = (x-a)(x-b)(x-c)$



What's the intuition behind this? Why is this true?




functions roots regular-expressions




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    It should be $f(x)=k(x-a)dots$ instead
    – nicomezi
    Jul 25 at 13:18










  • As gandalf61 explained below, this is a consequence of the factor theorem, which is a special case of the remainder theorem.
    – user577413
    Jul 25 at 13:45














up vote
1
down vote

favorite












Say $f(x)$ is a polynomial function with roots $a, , b, , c$ then this can be expressed as $f(x) = (x-a)(x-b)(x-c)$



What's the intuition behind this? Why is this true?




functions roots regular-expressions




share|cite|improve this question















  • 1




    It should be $f(x)=k(x-a)dots$ instead
    – nicomezi
    Jul 25 at 13:18










  • As gandalf61 explained below, this is a consequence of the factor theorem, which is a special case of the remainder theorem.
    – user577413
    Jul 25 at 13:45












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Say $f(x)$ is a polynomial function with roots $a, , b, , c$ then this can be expressed as $f(x) = (x-a)(x-b)(x-c)$



What's the intuition behind this? Why is this true?




functions roots regular-expressions




share|cite|improve this question











Say $f(x)$ is a polynomial function with roots $a, , b, , c$ then this can be expressed as $f(x) = (x-a)(x-b)(x-c)$



What's the intuition behind this? Why is this true?





functions roots regular-expressions





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 25 at 13:17









William

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753214







  • 1




    It should be $f(x)=k(x-a)dots$ instead
    – nicomezi
    Jul 25 at 13:18










  • As gandalf61 explained below, this is a consequence of the factor theorem, which is a special case of the remainder theorem.
    – user577413
    Jul 25 at 13:45












  • 1




    It should be $f(x)=k(x-a)dots$ instead
    – nicomezi
    Jul 25 at 13:18










  • As gandalf61 explained below, this is a consequence of the factor theorem, which is a special case of the remainder theorem.
    – user577413
    Jul 25 at 13:45







1




1




It should be $f(x)=k(x-a)dots$ instead
– nicomezi
Jul 25 at 13:18




It should be $f(x)=k(x-a)dots$ instead
– nicomezi
Jul 25 at 13:18












As gandalf61 explained below, this is a consequence of the factor theorem, which is a special case of the remainder theorem.
– user577413
Jul 25 at 13:45




As gandalf61 explained below, this is a consequence of the factor theorem, which is a special case of the remainder theorem.
– user577413
Jul 25 at 13:45










1 Answer
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The Euclidean algorithm applied to polynomials tells us that for any polynomial $f(x)$ then



$f(x) = (x-a)q(x) + r$



for some polynomial $q(x)$ and some constant $r$. Note that $r$ is a constant because it must have a degree strictly less than the degree of $x-a$, which is $1$.



But if $a$ is a root of $f(x)$ then $f(a)=0$, so $r=0$, and so



$f(x) = (x-a)q(x)$



i.e. $x-a$ is a factor of $f(x)$.






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    up vote
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    The Euclidean algorithm applied to polynomials tells us that for any polynomial $f(x)$ then



    $f(x) = (x-a)q(x) + r$



    for some polynomial $q(x)$ and some constant $r$. Note that $r$ is a constant because it must have a degree strictly less than the degree of $x-a$, which is $1$.



    But if $a$ is a root of $f(x)$ then $f(a)=0$, so $r=0$, and so



    $f(x) = (x-a)q(x)$



    i.e. $x-a$ is a factor of $f(x)$.






    share|cite|improve this answer

























      up vote
      6
      down vote













      The Euclidean algorithm applied to polynomials tells us that for any polynomial $f(x)$ then



      $f(x) = (x-a)q(x) + r$



      for some polynomial $q(x)$ and some constant $r$. Note that $r$ is a constant because it must have a degree strictly less than the degree of $x-a$, which is $1$.



      But if $a$ is a root of $f(x)$ then $f(a)=0$, so $r=0$, and so



      $f(x) = (x-a)q(x)$



      i.e. $x-a$ is a factor of $f(x)$.






      share|cite|improve this answer























        up vote
        6
        down vote










        up vote
        6
        down vote









        The Euclidean algorithm applied to polynomials tells us that for any polynomial $f(x)$ then



        $f(x) = (x-a)q(x) + r$



        for some polynomial $q(x)$ and some constant $r$. Note that $r$ is a constant because it must have a degree strictly less than the degree of $x-a$, which is $1$.



        But if $a$ is a root of $f(x)$ then $f(a)=0$, so $r=0$, and so



        $f(x) = (x-a)q(x)$



        i.e. $x-a$ is a factor of $f(x)$.






        share|cite|improve this answer













        The Euclidean algorithm applied to polynomials tells us that for any polynomial $f(x)$ then



        $f(x) = (x-a)q(x) + r$



        for some polynomial $q(x)$ and some constant $r$. Note that $r$ is a constant because it must have a degree strictly less than the degree of $x-a$, which is $1$.



        But if $a$ is a root of $f(x)$ then $f(a)=0$, so $r=0$, and so



        $f(x) = (x-a)q(x)$



        i.e. $x-a$ is a factor of $f(x)$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 25 at 13:23









        gandalf61

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