A Question On Remainder A Factor Theorem

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Let's take polynomial $p(x)$ and say that $p(x)/(x-3)=q(x)+2/(x-3)$.



Now let's say that we set the value of $x=3$. Does this prove that when $p(x)=3$, $p(x)=2$?



Part of me thinks it would, because $p(x)=q(x)(x-3) + 2$. And when $x=3$, we are left with $p(x)=2$.



But another part of me thinks that it wouldn't because if $x=3$:



$ - (x-3) * p(x)/(x-3) = (x-3) * q(x)+(x-3) * (2/(x-3))$
$ - 0 * p(x)/0 = 0+(x-3) * 0(2/0)$



So wouldn't the answer be undefined, because $0$ times $2/(x-3)$ when $x=3 = 0 $ times $2/0$ iss undefined?



So is the answer 2 or undefined? and why is the other interpretation incorrect?




algebra-precalculus functions polynomials




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  • When $x = 3$, we have $p(x) = 2$ at $x = 3$ only (i.e. $p(3) = 2$).
    – K_inverse
    Aug 3 at 8:10







  • 1




    Mathjax tip: use frac ab to get $frac ab$
    – Mohammad Zuhair Khan
    Aug 3 at 8:34














up vote
0
down vote

favorite
1












Let's take polynomial $p(x)$ and say that $p(x)/(x-3)=q(x)+2/(x-3)$.



Now let's say that we set the value of $x=3$. Does this prove that when $p(x)=3$, $p(x)=2$?



Part of me thinks it would, because $p(x)=q(x)(x-3) + 2$. And when $x=3$, we are left with $p(x)=2$.



But another part of me thinks that it wouldn't because if $x=3$:



$ - (x-3) * p(x)/(x-3) = (x-3) * q(x)+(x-3) * (2/(x-3))$
$ - 0 * p(x)/0 = 0+(x-3) * 0(2/0)$



So wouldn't the answer be undefined, because $0$ times $2/(x-3)$ when $x=3 = 0 $ times $2/0$ iss undefined?



So is the answer 2 or undefined? and why is the other interpretation incorrect?




algebra-precalculus functions polynomials




share|cite|improve this question





















  • When $x = 3$, we have $p(x) = 2$ at $x = 3$ only (i.e. $p(3) = 2$).
    – K_inverse
    Aug 3 at 8:10







  • 1




    Mathjax tip: use frac ab to get $frac ab$
    – Mohammad Zuhair Khan
    Aug 3 at 8:34












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





Let's take polynomial $p(x)$ and say that $p(x)/(x-3)=q(x)+2/(x-3)$.



Now let's say that we set the value of $x=3$. Does this prove that when $p(x)=3$, $p(x)=2$?



Part of me thinks it would, because $p(x)=q(x)(x-3) + 2$. And when $x=3$, we are left with $p(x)=2$.



But another part of me thinks that it wouldn't because if $x=3$:



$ - (x-3) * p(x)/(x-3) = (x-3) * q(x)+(x-3) * (2/(x-3))$
$ - 0 * p(x)/0 = 0+(x-3) * 0(2/0)$



So wouldn't the answer be undefined, because $0$ times $2/(x-3)$ when $x=3 = 0 $ times $2/0$ iss undefined?



So is the answer 2 or undefined? and why is the other interpretation incorrect?




algebra-precalculus functions polynomials




share|cite|improve this question













Let's take polynomial $p(x)$ and say that $p(x)/(x-3)=q(x)+2/(x-3)$.



Now let's say that we set the value of $x=3$. Does this prove that when $p(x)=3$, $p(x)=2$?



Part of me thinks it would, because $p(x)=q(x)(x-3) + 2$. And when $x=3$, we are left with $p(x)=2$.



But another part of me thinks that it wouldn't because if $x=3$:



$ - (x-3) * p(x)/(x-3) = (x-3) * q(x)+(x-3) * (2/(x-3))$
$ - 0 * p(x)/0 = 0+(x-3) * 0(2/0)$



So wouldn't the answer be undefined, because $0$ times $2/(x-3)$ when $x=3 = 0 $ times $2/0$ iss undefined?



So is the answer 2 or undefined? and why is the other interpretation incorrect?





algebra-precalculus functions polynomials





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 3 at 8:31









pointguard0

605417




605417









asked Aug 3 at 8:06









Ethan Chan

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591322











  • When $x = 3$, we have $p(x) = 2$ at $x = 3$ only (i.e. $p(3) = 2$).
    – K_inverse
    Aug 3 at 8:10







  • 1




    Mathjax tip: use frac ab to get $frac ab$
    – Mohammad Zuhair Khan
    Aug 3 at 8:34
















  • When $x = 3$, we have $p(x) = 2$ at $x = 3$ only (i.e. $p(3) = 2$).
    – K_inverse
    Aug 3 at 8:10







  • 1




    Mathjax tip: use frac ab to get $frac ab$
    – Mohammad Zuhair Khan
    Aug 3 at 8:34















When $x = 3$, we have $p(x) = 2$ at $x = 3$ only (i.e. $p(3) = 2$).
– K_inverse
Aug 3 at 8:10





When $x = 3$, we have $p(x) = 2$ at $x = 3$ only (i.e. $p(3) = 2$).
– K_inverse
Aug 3 at 8:10





1




1




Mathjax tip: use frac ab to get $frac ab$
– Mohammad Zuhair Khan
Aug 3 at 8:34




Mathjax tip: use frac ab to get $frac ab$
– Mohammad Zuhair Khan
Aug 3 at 8:34










1 Answer
1






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Do not divide by zero.



Writing $fracp(x)(x-3)$ means you are assuming the case when $x ne 3$.



If we indeed have $forall xin mathbbR, p(x)=q(x)(x-3)+2$ where $q$ is a polynomial, then $p(3)=2$.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

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    active

    oldest

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    up vote
    1
    down vote



    accepted










    Do not divide by zero.



    Writing $fracp(x)(x-3)$ means you are assuming the case when $x ne 3$.



    If we indeed have $forall xin mathbbR, p(x)=q(x)(x-3)+2$ where $q$ is a polynomial, then $p(3)=2$.






    share|cite|improve this answer



























      up vote
      1
      down vote



      accepted










      Do not divide by zero.



      Writing $fracp(x)(x-3)$ means you are assuming the case when $x ne 3$.



      If we indeed have $forall xin mathbbR, p(x)=q(x)(x-3)+2$ where $q$ is a polynomial, then $p(3)=2$.






      share|cite|improve this answer

























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        Do not divide by zero.



        Writing $fracp(x)(x-3)$ means you are assuming the case when $x ne 3$.



        If we indeed have $forall xin mathbbR, p(x)=q(x)(x-3)+2$ where $q$ is a polynomial, then $p(3)=2$.






        share|cite|improve this answer















        Do not divide by zero.



        Writing $fracp(x)(x-3)$ means you are assuming the case when $x ne 3$.



        If we indeed have $forall xin mathbbR, p(x)=q(x)(x-3)+2$ where $q$ is a polynomial, then $p(3)=2$.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 3 at 8:24


























        answered Aug 3 at 8:12









        Siong Thye Goh

        76.6k134794




        76.6k134794






















             

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