Method to factor an expression












1












$begingroup$


As the title says, i want to factorize an expression, but i don't have any clue how to proceed.



Here is the expression :



$2x² -7x +3$



And here is the factorized form :



$(x-3)(2x - 1)$



My question is, which method or rule to use to go from first to second ?



please note that I am a beginner, and the only question that i found which is closer to mine is this post.



Thank you for your help !






factoring







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    As the title says, i want to factorize an expression, but i don't have any clue how to proceed.



    Here is the expression :



    $2x² -7x +3$



    And here is the factorized form :



    $(x-3)(2x - 1)$



    My question is, which method or rule to use to go from first to second ?



    please note that I am a beginner, and the only question that i found which is closer to mine is this post.



    Thank you for your help !






    factoring







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      As the title says, i want to factorize an expression, but i don't have any clue how to proceed.



      Here is the expression :



      $2x² -7x +3$



      And here is the factorized form :



      $(x-3)(2x - 1)$



      My question is, which method or rule to use to go from first to second ?



      please note that I am a beginner, and the only question that i found which is closer to mine is this post.



      Thank you for your help !






      factoring







      share|cite|improve this question









      $endgroup$




      As the title says, i want to factorize an expression, but i don't have any clue how to proceed.



      Here is the expression :



      $2x² -7x +3$



      And here is the factorized form :



      $(x-3)(2x - 1)$



      My question is, which method or rule to use to go from first to second ?



      please note that I am a beginner, and the only question that i found which is closer to mine is this post.



      Thank you for your help !






      factoring




      factoring






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 18 '18 at 9:18









      ganzo dbganzo db

      155




      155






















          4 Answers
          4






          active

          oldest

          votes


















          3












          $begingroup$

          Rewrite the expression into the form:



          $2x^2-6x-x+3$ ,



          then group the first two terms together and the last two terms together:



          $(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$ ,



          next extract the common factor:



          $(x-3)(2x-1)$ ,



          and you are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I like the simplicity of your method, thank's :)
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:41










          • $begingroup$
            @ganzodb You are welcome
            $endgroup$
            – Matko
            Dec 18 '18 at 18:24



















          2












          $begingroup$

          If it concerns quadratic polynomial $ax^2+bx+c$ then start with calculating discriminant: $$D:=b^2-4ac$$



          If $D$ is negative then give up (unless you are familiar with complex numbers already).



          If $D$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $x_1=frac{-b+sqrt D}{2a}$ and $x_2=frac{-b-sqrt D}{2a}$.



          Especially if $D$ is a perfect square (as in your case, where $D=25$) then there is reason to cheer.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            As you supposed, i'm not good with complex numbers :) !
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:54



















          2












          $begingroup$

          It relies on the following property of quadratic polynomials:




          If the quadratic polynomial $;ax^2+bx+c;(ane 0)$ has roots $xi_0$ and $xi_1$ (real or complex, distinct ot not), it can factored as
          $$ax^2+bx+c=a(x-xi_0)(x-xi_1).$$




          This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:




          If a polynomial $p(x)$ has root $xi$, it is divisible by $x-xi$.




          In the present case, the discriminat of $2x² -7x +3$ is $;Delta=49-4cdot 2cdot 3=25$, so its roots are $;frac{7pm 5}4==bigl{3,frac 12bigr}$, and the factorisation is
          $$2(x-3)Bigl(x-frac12Bigr)=(x-3)(2x-1).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i wish i could understand the last one ( frac{7pm 5}4 ).
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:57








          • 2




            $begingroup$
            @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
            $endgroup$
            – Bernard
            Dec 18 '18 at 18:22



















          2












          $begingroup$

          In general ,if $x_1$ and $x_2$ are roots of $$underbrace{a}_{neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$text{sum of the roots}=-frac{b}{a};;&;;text{product of the roots}=frac{c}{a}$$




          So your case, $x_1+x_2= frac{7}{2}$ and $x_1x_2=frac{3}{2}$



          Now solve these to get $x_1$ and $x_2$ to finish your conclusion




          To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-frac{7}{2}x+frac{3}{2}=0$$ which means $$x^2-2left(frac{7}{4}right)x=-frac{3}{2}$$ which is same as $$x^2-2left(frac{7}{4}right)x+frac{49}{16}=-frac{3}{2}+frac{49}{16}=frac{25}{16}$$ so $$left(x-frac{7}{4}right)^2=frac{25}{16}$$ and so $$x-frac{7}{4}=pm sqrt{frac{25}{16}}=pm frac{5}{4}$$ so $$x=frac{7}{4} pm frac{5}{4}=frac{7pm 5}{4}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:53











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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Rewrite the expression into the form:



          $2x^2-6x-x+3$ ,



          then group the first two terms together and the last two terms together:



          $(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$ ,



          next extract the common factor:



          $(x-3)(2x-1)$ ,



          and you are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I like the simplicity of your method, thank's :)
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:41










          • $begingroup$
            @ganzodb You are welcome
            $endgroup$
            – Matko
            Dec 18 '18 at 18:24
















          3












          $begingroup$

          Rewrite the expression into the form:



          $2x^2-6x-x+3$ ,



          then group the first two terms together and the last two terms together:



          $(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$ ,



          next extract the common factor:



          $(x-3)(2x-1)$ ,



          and you are done.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I like the simplicity of your method, thank's :)
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:41










          • $begingroup$
            @ganzodb You are welcome
            $endgroup$
            – Matko
            Dec 18 '18 at 18:24














          3












          3








          3





          $begingroup$

          Rewrite the expression into the form:



          $2x^2-6x-x+3$ ,



          then group the first two terms together and the last two terms together:



          $(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$ ,



          next extract the common factor:



          $(x-3)(2x-1)$ ,



          and you are done.






          share|cite|improve this answer











          $endgroup$



          Rewrite the expression into the form:



          $2x^2-6x-x+3$ ,



          then group the first two terms together and the last two terms together:



          $(2x^2-6x)-(x-3)=2x(x-3)-(x-3)$ ,



          next extract the common factor:



          $(x-3)(2x-1)$ ,



          and you are done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 10:11

























          answered Dec 18 '18 at 10:05









          MatkoMatko

          864




          864












          • $begingroup$
            I like the simplicity of your method, thank's :)
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:41










          • $begingroup$
            @ganzodb You are welcome
            $endgroup$
            – Matko
            Dec 18 '18 at 18:24


















          • $begingroup$
            I like the simplicity of your method, thank's :)
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:41










          • $begingroup$
            @ganzodb You are welcome
            $endgroup$
            – Matko
            Dec 18 '18 at 18:24
















          $begingroup$
          I like the simplicity of your method, thank's :)
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:41




          $begingroup$
          I like the simplicity of your method, thank's :)
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:41












          $begingroup$
          @ganzodb You are welcome
          $endgroup$
          – Matko
          Dec 18 '18 at 18:24




          $begingroup$
          @ganzodb You are welcome
          $endgroup$
          – Matko
          Dec 18 '18 at 18:24











          2












          $begingroup$

          If it concerns quadratic polynomial $ax^2+bx+c$ then start with calculating discriminant: $$D:=b^2-4ac$$



          If $D$ is negative then give up (unless you are familiar with complex numbers already).



          If $D$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $x_1=frac{-b+sqrt D}{2a}$ and $x_2=frac{-b-sqrt D}{2a}$.



          Especially if $D$ is a perfect square (as in your case, where $D=25$) then there is reason to cheer.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            As you supposed, i'm not good with complex numbers :) !
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:54
















          2












          $begingroup$

          If it concerns quadratic polynomial $ax^2+bx+c$ then start with calculating discriminant: $$D:=b^2-4ac$$



          If $D$ is negative then give up (unless you are familiar with complex numbers already).



          If $D$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $x_1=frac{-b+sqrt D}{2a}$ and $x_2=frac{-b-sqrt D}{2a}$.



          Especially if $D$ is a perfect square (as in your case, where $D=25$) then there is reason to cheer.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            As you supposed, i'm not good with complex numbers :) !
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:54














          2












          2








          2





          $begingroup$

          If it concerns quadratic polynomial $ax^2+bx+c$ then start with calculating discriminant: $$D:=b^2-4ac$$



          If $D$ is negative then give up (unless you are familiar with complex numbers already).



          If $D$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $x_1=frac{-b+sqrt D}{2a}$ and $x_2=frac{-b-sqrt D}{2a}$.



          Especially if $D$ is a perfect square (as in your case, where $D=25$) then there is reason to cheer.






          share|cite|improve this answer









          $endgroup$



          If it concerns quadratic polynomial $ax^2+bx+c$ then start with calculating discriminant: $$D:=b^2-4ac$$



          If $D$ is negative then give up (unless you are familiar with complex numbers already).



          If $D$ is nonnegative then: $$ax^2+bx+c=a(x-x_1)(x-x_2)$$ where $x_1=frac{-b+sqrt D}{2a}$ and $x_2=frac{-b-sqrt D}{2a}$.



          Especially if $D$ is a perfect square (as in your case, where $D=25$) then there is reason to cheer.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 9:39









          drhabdrhab

          101k545136




          101k545136












          • $begingroup$
            As you supposed, i'm not good with complex numbers :) !
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:54


















          • $begingroup$
            As you supposed, i'm not good with complex numbers :) !
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:54
















          $begingroup$
          As you supposed, i'm not good with complex numbers :) !
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:54




          $begingroup$
          As you supposed, i'm not good with complex numbers :) !
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:54











          2












          $begingroup$

          It relies on the following property of quadratic polynomials:




          If the quadratic polynomial $;ax^2+bx+c;(ane 0)$ has roots $xi_0$ and $xi_1$ (real or complex, distinct ot not), it can factored as
          $$ax^2+bx+c=a(x-xi_0)(x-xi_1).$$




          This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:




          If a polynomial $p(x)$ has root $xi$, it is divisible by $x-xi$.




          In the present case, the discriminat of $2x² -7x +3$ is $;Delta=49-4cdot 2cdot 3=25$, so its roots are $;frac{7pm 5}4==bigl{3,frac 12bigr}$, and the factorisation is
          $$2(x-3)Bigl(x-frac12Bigr)=(x-3)(2x-1).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i wish i could understand the last one ( frac{7pm 5}4 ).
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:57








          • 2




            $begingroup$
            @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
            $endgroup$
            – Bernard
            Dec 18 '18 at 18:22
















          2












          $begingroup$

          It relies on the following property of quadratic polynomials:




          If the quadratic polynomial $;ax^2+bx+c;(ane 0)$ has roots $xi_0$ and $xi_1$ (real or complex, distinct ot not), it can factored as
          $$ax^2+bx+c=a(x-xi_0)(x-xi_1).$$




          This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:




          If a polynomial $p(x)$ has root $xi$, it is divisible by $x-xi$.




          In the present case, the discriminat of $2x² -7x +3$ is $;Delta=49-4cdot 2cdot 3=25$, so its roots are $;frac{7pm 5}4==bigl{3,frac 12bigr}$, and the factorisation is
          $$2(x-3)Bigl(x-frac12Bigr)=(x-3)(2x-1).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            i wish i could understand the last one ( frac{7pm 5}4 ).
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:57








          • 2




            $begingroup$
            @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
            $endgroup$
            – Bernard
            Dec 18 '18 at 18:22














          2












          2








          2





          $begingroup$

          It relies on the following property of quadratic polynomials:




          If the quadratic polynomial $;ax^2+bx+c;(ane 0)$ has roots $xi_0$ and $xi_1$ (real or complex, distinct ot not), it can factored as
          $$ax^2+bx+c=a(x-xi_0)(x-xi_1).$$




          This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:




          If a polynomial $p(x)$ has root $xi$, it is divisible by $x-xi$.




          In the present case, the discriminat of $2x² -7x +3$ is $;Delta=49-4cdot 2cdot 3=25$, so its roots are $;frac{7pm 5}4==bigl{3,frac 12bigr}$, and the factorisation is
          $$2(x-3)Bigl(x-frac12Bigr)=(x-3)(2x-1).$$






          share|cite|improve this answer









          $endgroup$



          It relies on the following property of quadratic polynomials:




          If the quadratic polynomial $;ax^2+bx+c;(ane 0)$ has roots $xi_0$ and $xi_1$ (real or complex, distinct ot not), it can factored as
          $$ax^2+bx+c=a(x-xi_0)(x-xi_1).$$




          This property is a consequence of a more general property of polynomials (of any degree) and the ring of polynomials over a field being a euclidean domain:




          If a polynomial $p(x)$ has root $xi$, it is divisible by $x-xi$.




          In the present case, the discriminat of $2x² -7x +3$ is $;Delta=49-4cdot 2cdot 3=25$, so its roots are $;frac{7pm 5}4==bigl{3,frac 12bigr}$, and the factorisation is
          $$2(x-3)Bigl(x-frac12Bigr)=(x-3)(2x-1).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 9:40









          BernardBernard

          121k740116




          121k740116












          • $begingroup$
            i wish i could understand the last one ( frac{7pm 5}4 ).
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:57








          • 2




            $begingroup$
            @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
            $endgroup$
            – Bernard
            Dec 18 '18 at 18:22


















          • $begingroup$
            i wish i could understand the last one ( frac{7pm 5}4 ).
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:57








          • 2




            $begingroup$
            @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
            $endgroup$
            – Bernard
            Dec 18 '18 at 18:22
















          $begingroup$
          i wish i could understand the last one ( frac{7pm 5}4 ).
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:57






          $begingroup$
          i wish i could understand the last one ( frac{7pm 5}4 ).
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:57






          2




          2




          $begingroup$
          @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
          $endgroup$
          – Bernard
          Dec 18 '18 at 18:22




          $begingroup$
          @ganzodb: It's just the formula $frac{-bpmsqrtDelta}{2a}.$.
          $endgroup$
          – Bernard
          Dec 18 '18 at 18:22











          2












          $begingroup$

          In general ,if $x_1$ and $x_2$ are roots of $$underbrace{a}_{neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$text{sum of the roots}=-frac{b}{a};;&;;text{product of the roots}=frac{c}{a}$$




          So your case, $x_1+x_2= frac{7}{2}$ and $x_1x_2=frac{3}{2}$



          Now solve these to get $x_1$ and $x_2$ to finish your conclusion




          To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-frac{7}{2}x+frac{3}{2}=0$$ which means $$x^2-2left(frac{7}{4}right)x=-frac{3}{2}$$ which is same as $$x^2-2left(frac{7}{4}right)x+frac{49}{16}=-frac{3}{2}+frac{49}{16}=frac{25}{16}$$ so $$left(x-frac{7}{4}right)^2=frac{25}{16}$$ and so $$x-frac{7}{4}=pm sqrt{frac{25}{16}}=pm frac{5}{4}$$ so $$x=frac{7}{4} pm frac{5}{4}=frac{7pm 5}{4}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:53
















          2












          $begingroup$

          In general ,if $x_1$ and $x_2$ are roots of $$underbrace{a}_{neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$text{sum of the roots}=-frac{b}{a};;&;;text{product of the roots}=frac{c}{a}$$




          So your case, $x_1+x_2= frac{7}{2}$ and $x_1x_2=frac{3}{2}$



          Now solve these to get $x_1$ and $x_2$ to finish your conclusion




          To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-frac{7}{2}x+frac{3}{2}=0$$ which means $$x^2-2left(frac{7}{4}right)x=-frac{3}{2}$$ which is same as $$x^2-2left(frac{7}{4}right)x+frac{49}{16}=-frac{3}{2}+frac{49}{16}=frac{25}{16}$$ so $$left(x-frac{7}{4}right)^2=frac{25}{16}$$ and so $$x-frac{7}{4}=pm sqrt{frac{25}{16}}=pm frac{5}{4}$$ so $$x=frac{7}{4} pm frac{5}{4}=frac{7pm 5}{4}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:53














          2












          2








          2





          $begingroup$

          In general ,if $x_1$ and $x_2$ are roots of $$underbrace{a}_{neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$text{sum of the roots}=-frac{b}{a};;&;;text{product of the roots}=frac{c}{a}$$




          So your case, $x_1+x_2= frac{7}{2}$ and $x_1x_2=frac{3}{2}$



          Now solve these to get $x_1$ and $x_2$ to finish your conclusion




          To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-frac{7}{2}x+frac{3}{2}=0$$ which means $$x^2-2left(frac{7}{4}right)x=-frac{3}{2}$$ which is same as $$x^2-2left(frac{7}{4}right)x+frac{49}{16}=-frac{3}{2}+frac{49}{16}=frac{25}{16}$$ so $$left(x-frac{7}{4}right)^2=frac{25}{16}$$ and so $$x-frac{7}{4}=pm sqrt{frac{25}{16}}=pm frac{5}{4}$$ so $$x=frac{7}{4} pm frac{5}{4}=frac{7pm 5}{4}$$






          share|cite|improve this answer











          $endgroup$



          In general ,if $x_1$ and $x_2$ are roots of $$underbrace{a}_{neq 0}x^2+bx+c=0$$ then $$ax^2+bx+c=k(x-x_1)(x-x_2)=k[x^2-(x_1+x_2)x+x_1x_2]$$ comparing the coefficients, $$a=k,b=-k(x_1+x_2),c=kx_1x_2$$ Consequently $$text{sum of the roots}=-frac{b}{a};;&;;text{product of the roots}=frac{c}{a}$$




          So your case, $x_1+x_2= frac{7}{2}$ and $x_1x_2=frac{3}{2}$



          Now solve these to get $x_1$ and $x_2$ to finish your conclusion




          To find $x_1$ and $x_2$, $$2x^2-7x+3=0$$ implies $$x^2-frac{7}{2}x+frac{3}{2}=0$$ which means $$x^2-2left(frac{7}{4}right)x=-frac{3}{2}$$ which is same as $$x^2-2left(frac{7}{4}right)x+frac{49}{16}=-frac{3}{2}+frac{49}{16}=frac{25}{16}$$ so $$left(x-frac{7}{4}right)^2=frac{25}{16}$$ and so $$x-frac{7}{4}=pm sqrt{frac{25}{16}}=pm frac{5}{4}$$ so $$x=frac{7}{4} pm frac{5}{4}=frac{7pm 5}{4}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 10:17

























          answered Dec 18 '18 at 10:04









          Chinnapparaj RChinnapparaj R

          5,4872928




          5,4872928












          • $begingroup$
            Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:53


















          • $begingroup$
            Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
            $endgroup$
            – ganzo db
            Dec 18 '18 at 16:53
















          $begingroup$
          Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:53




          $begingroup$
          Thank you for the time spent in the explanation, i'm still unfamiliar with some mathematical concepts.
          $endgroup$
          – ganzo db
          Dec 18 '18 at 16:53


















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