Factorise polynomial with real and complex roots












7












$begingroup$


How would you go about finding the roots of the polynomial:
$$x^4 +5x^3+4x^2+6x-4=0.$$



I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations.



I was wondering if there was a better approach to this question.



Any help would be greatly appreciated.






polynomials complex-numbers factoring quartic-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Rational Root Test? See: example: math.stackexchange.com/questions/12787/…
    $endgroup$
    – Moo
    Jan 31 at 22:20












  • $begingroup$
    Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution.
    $endgroup$
    – Peter
    Jan 31 at 22:20










  • $begingroup$
    Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$.
    $endgroup$
    – Peter
    Jan 31 at 22:22


















7












$begingroup$


How would you go about finding the roots of the polynomial:
$$x^4 +5x^3+4x^2+6x-4=0.$$



I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations.



I was wondering if there was a better approach to this question.



Any help would be greatly appreciated.






polynomials complex-numbers factoring quartic-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Rational Root Test? See: example: math.stackexchange.com/questions/12787/…
    $endgroup$
    – Moo
    Jan 31 at 22:20












  • $begingroup$
    Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution.
    $endgroup$
    – Peter
    Jan 31 at 22:20










  • $begingroup$
    Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$.
    $endgroup$
    – Peter
    Jan 31 at 22:22
















7












7








7


1



$begingroup$


How would you go about finding the roots of the polynomial:
$$x^4 +5x^3+4x^2+6x-4=0.$$



I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations.



I was wondering if there was a better approach to this question.



Any help would be greatly appreciated.






polynomials complex-numbers factoring quartic-equations







share|cite|improve this question











$endgroup$




How would you go about finding the roots of the polynomial:
$$x^4 +5x^3+4x^2+6x-4=0.$$



I attempt to form two quadratics e.g.
$$(x^2+ax+b)(x^2+cx+d)$$
and then tried to expand, collect like terms, and solve for the coefficients simultaneously, but this results in a hard system of equations.



I was wondering if there was a better approach to this question.



Any help would be greatly appreciated.






polynomials complex-numbers factoring quartic-equations




polynomials complex-numbers factoring quartic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 31 at 23:08









Michael Rozenberg

108k1895200




108k1895200










asked Jan 31 at 22:16









ultralightultralight

786




786








  • 1




    $begingroup$
    Rational Root Test? See: example: math.stackexchange.com/questions/12787/…
    $endgroup$
    – Moo
    Jan 31 at 22:20












  • $begingroup$
    Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution.
    $endgroup$
    – Peter
    Jan 31 at 22:20










  • $begingroup$
    Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$.
    $endgroup$
    – Peter
    Jan 31 at 22:22
















  • 1




    $begingroup$
    Rational Root Test? See: example: math.stackexchange.com/questions/12787/…
    $endgroup$
    – Moo
    Jan 31 at 22:20












  • $begingroup$
    Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution.
    $endgroup$
    – Peter
    Jan 31 at 22:20










  • $begingroup$
    Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$.
    $endgroup$
    – Peter
    Jan 31 at 22:22










1




1




$begingroup$
Rational Root Test? See: example: math.stackexchange.com/questions/12787/…
$endgroup$
– Moo
Jan 31 at 22:20






$begingroup$
Rational Root Test? See: example: math.stackexchange.com/questions/12787/…
$endgroup$
– Moo
Jan 31 at 22:20














$begingroup$
Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution.
$endgroup$
– Peter
Jan 31 at 22:20




$begingroup$
Not sure whether Kronecker's method or factoring modulo some primes gives a faster solution.
$endgroup$
– Peter
Jan 31 at 22:20












$begingroup$
Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$.
$endgroup$
– Peter
Jan 31 at 22:22






$begingroup$
Maybe you try all the possible pairs of $b$ and $d$ multiplying to $-4$ ($b$ and $d$ must be integers). For every such pair, try to find $a$ and $c$.
$endgroup$
– Peter
Jan 31 at 22:22












3 Answers
3






active

oldest

votes


















6












$begingroup$

Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
    $endgroup$
    – Robert Z
    Jan 31 at 22:36








  • 2




    $begingroup$
    Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
    $endgroup$
    – José Carlos Santos
    Jan 31 at 22:43






  • 2




    $begingroup$
    Really? I think you were very lucky with your first approach.
    $endgroup$
    – Robert Z
    Jan 31 at 22:46






  • 1




    $begingroup$
    Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
    $endgroup$
    – drjpizzle
    Feb 1 at 1:32






  • 2




    $begingroup$
    I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
    $endgroup$
    – José Carlos Santos
    Feb 1 at 7:16



















4












$begingroup$

Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as
$$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$
is a good idea.



Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples:
$$(b,d)in{(4,-1), (2,-2), (-4,1)}$$
and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$.



P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:



i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.



ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    I like the following way.



    For all real value of $k$ we have:



    $$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}x+kright)^2-left(left(2k+frac{9}{4}right)x^2+(10k-6)x+k^2+4right).$$
    Now, we'll choose a value of $k$, for which $2k+frac{9}{4}>0$ and $$(5k-3)^2-left(2k+frac{9}{4}right)(k^2+4)=0.$$
    Easy to see that $k=0$ is valid.



    Thus,$$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}xright)^2-left(frac{9}{4}x^2-6x+4right)=$$
    $$=left(x^2+frac{5}{2}xright)^2-left(frac{3}{2}x-2right)^2=(x^2+x+2)(x^2+4x-2).$$
    Can you end it now?






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
        $endgroup$
        – Robert Z
        Jan 31 at 22:36








      • 2




        $begingroup$
        Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
        $endgroup$
        – José Carlos Santos
        Jan 31 at 22:43






      • 2




        $begingroup$
        Really? I think you were very lucky with your first approach.
        $endgroup$
        – Robert Z
        Jan 31 at 22:46






      • 1




        $begingroup$
        Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
        $endgroup$
        – drjpizzle
        Feb 1 at 1:32






      • 2




        $begingroup$
        I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
        $endgroup$
        – José Carlos Santos
        Feb 1 at 7:16
















      6












      $begingroup$

      Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
        $endgroup$
        – Robert Z
        Jan 31 at 22:36








      • 2




        $begingroup$
        Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
        $endgroup$
        – José Carlos Santos
        Jan 31 at 22:43






      • 2




        $begingroup$
        Really? I think you were very lucky with your first approach.
        $endgroup$
        – Robert Z
        Jan 31 at 22:46






      • 1




        $begingroup$
        Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
        $endgroup$
        – drjpizzle
        Feb 1 at 1:32






      • 2




        $begingroup$
        I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
        $endgroup$
        – José Carlos Santos
        Feb 1 at 7:16














      6












      6








      6





      $begingroup$

      Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.






      share|cite|improve this answer









      $endgroup$



      Note that $x^4-4=(x^2+2)(x^2-2)$. This suggests that you might try to express your polynomial as$$(x^2+ax+2)(x^2+bx-2)=x^4+(a+b)x^3+abx^2+2(-a+b)x-4.$$It is now easy to see that all it takes is to choose $a=1$ and $b=4$.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 31 at 22:29









      José Carlos SantosJosé Carlos Santos

      168k22132236




      168k22132236








      • 2




        $begingroup$
        I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
        $endgroup$
        – Robert Z
        Jan 31 at 22:36








      • 2




        $begingroup$
        Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
        $endgroup$
        – José Carlos Santos
        Jan 31 at 22:43






      • 2




        $begingroup$
        Really? I think you were very lucky with your first approach.
        $endgroup$
        – Robert Z
        Jan 31 at 22:46






      • 1




        $begingroup$
        Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
        $endgroup$
        – drjpizzle
        Feb 1 at 1:32






      • 2




        $begingroup$
        I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
        $endgroup$
        – José Carlos Santos
        Feb 1 at 7:16














      • 2




        $begingroup$
        I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
        $endgroup$
        – Robert Z
        Jan 31 at 22:36








      • 2




        $begingroup$
        Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
        $endgroup$
        – José Carlos Santos
        Jan 31 at 22:43






      • 2




        $begingroup$
        Really? I think you were very lucky with your first approach.
        $endgroup$
        – Robert Z
        Jan 31 at 22:46






      • 1




        $begingroup$
        Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
        $endgroup$
        – drjpizzle
        Feb 1 at 1:32






      • 2




        $begingroup$
        I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
        $endgroup$
        – José Carlos Santos
        Feb 1 at 7:16








      2




      2




      $begingroup$
      I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
      $endgroup$
      – Robert Z
      Jan 31 at 22:36






      $begingroup$
      I don't understand the initial remark. If you have $x^4+2x^2-5x-4$ then your suggestion is not very useful because the factorization is $(x^2+x+4)(x^2-x-1)$.
      $endgroup$
      – Robert Z
      Jan 31 at 22:36






      2




      2




      $begingroup$
      Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
      $endgroup$
      – José Carlos Santos
      Jan 31 at 22:43




      $begingroup$
      Indeed. But what I wrote here corresponds to the way I approached the problem. In this case, it worked. If it had not worked, then I might have tried the general method.
      $endgroup$
      – José Carlos Santos
      Jan 31 at 22:43




      2




      2




      $begingroup$
      Really? I think you were very lucky with your first approach.
      $endgroup$
      – Robert Z
      Jan 31 at 22:46




      $begingroup$
      Really? I think you were very lucky with your first approach.
      $endgroup$
      – Robert Z
      Jan 31 at 22:46




      1




      1




      $begingroup$
      Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
      $endgroup$
      – drjpizzle
      Feb 1 at 1:32




      $begingroup$
      Does this answer boil down to: "Assume there are nice integer coefficients somewhere: that make things easier to brute force."? In general, this is not so simple a task.
      $endgroup$
      – drjpizzle
      Feb 1 at 1:32




      2




      2




      $begingroup$
      I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
      $endgroup$
      – José Carlos Santos
      Feb 1 at 7:16




      $begingroup$
      I tried it this way (after trying to find a rational root) and in this specific case it worked. It's not deeper than that.
      $endgroup$
      – José Carlos Santos
      Feb 1 at 7:16











      4












      $begingroup$

      Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as
      $$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$
      is a good idea.



      Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples:
      $$(b,d)in{(4,-1), (2,-2), (-4,1)}$$
      and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$.



      P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:



      i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.



      ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as
        $$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$
        is a good idea.



        Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples:
        $$(b,d)in{(4,-1), (2,-2), (-4,1)}$$
        and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$.



        P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:



        i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.



        ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as
          $$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$
          is a good idea.



          Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples:
          $$(b,d)in{(4,-1), (2,-2), (-4,1)}$$
          and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$.



          P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:



          i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.



          ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.






          share|cite|improve this answer











          $endgroup$



          Since there are no rational roots (a rational root should be a divisor of $4$), your attempt of factorization as
          $$x^4 +5x^3+4x^2+6x-4=(x^2+ax+b)(x^2+cx+d)$$
          is a good idea.



          Hint. By expanding the right-hand side and by comparing it with the the left-hand side, we have that $bd=-4$. Assuming that $b$ and $d$ are integers, by symmetry, you can try the following couples:
          $$(b,d)in{(4,-1), (2,-2), (-4,1)}$$
          and then solve the corresponding systems with respect to the remaining coefficients $c$ and $d$.



          P.S. If you are lucky you will start with the couple $(2,-2)$. Be careful though, there are polynomials, which are "similar" to the given one, such that the other couples work:



          i) $(4,-1)$ for $x^4+2x^2-5x-4=(x^2+x+4)(x^2-x-1)$.



          ii) $(-4,1)$ for $x^4-4x^2+5x-4=(x^2+x-4)(x^2-x+1)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Feb 1 at 11:33

























          answered Jan 31 at 22:25









          Robert ZRobert Z

          101k1069142




          101k1069142























              3












              $begingroup$

              I like the following way.



              For all real value of $k$ we have:



              $$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}x+kright)^2-left(left(2k+frac{9}{4}right)x^2+(10k-6)x+k^2+4right).$$
              Now, we'll choose a value of $k$, for which $2k+frac{9}{4}>0$ and $$(5k-3)^2-left(2k+frac{9}{4}right)(k^2+4)=0.$$
              Easy to see that $k=0$ is valid.



              Thus,$$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}xright)^2-left(frac{9}{4}x^2-6x+4right)=$$
              $$=left(x^2+frac{5}{2}xright)^2-left(frac{3}{2}x-2right)^2=(x^2+x+2)(x^2+4x-2).$$
              Can you end it now?






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                I like the following way.



                For all real value of $k$ we have:



                $$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}x+kright)^2-left(left(2k+frac{9}{4}right)x^2+(10k-6)x+k^2+4right).$$
                Now, we'll choose a value of $k$, for which $2k+frac{9}{4}>0$ and $$(5k-3)^2-left(2k+frac{9}{4}right)(k^2+4)=0.$$
                Easy to see that $k=0$ is valid.



                Thus,$$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}xright)^2-left(frac{9}{4}x^2-6x+4right)=$$
                $$=left(x^2+frac{5}{2}xright)^2-left(frac{3}{2}x-2right)^2=(x^2+x+2)(x^2+4x-2).$$
                Can you end it now?






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  I like the following way.



                  For all real value of $k$ we have:



                  $$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}x+kright)^2-left(left(2k+frac{9}{4}right)x^2+(10k-6)x+k^2+4right).$$
                  Now, we'll choose a value of $k$, for which $2k+frac{9}{4}>0$ and $$(5k-3)^2-left(2k+frac{9}{4}right)(k^2+4)=0.$$
                  Easy to see that $k=0$ is valid.



                  Thus,$$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}xright)^2-left(frac{9}{4}x^2-6x+4right)=$$
                  $$=left(x^2+frac{5}{2}xright)^2-left(frac{3}{2}x-2right)^2=(x^2+x+2)(x^2+4x-2).$$
                  Can you end it now?






                  share|cite|improve this answer









                  $endgroup$



                  I like the following way.



                  For all real value of $k$ we have:



                  $$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}x+kright)^2-left(left(2k+frac{9}{4}right)x^2+(10k-6)x+k^2+4right).$$
                  Now, we'll choose a value of $k$, for which $2k+frac{9}{4}>0$ and $$(5k-3)^2-left(2k+frac{9}{4}right)(k^2+4)=0.$$
                  Easy to see that $k=0$ is valid.



                  Thus,$$x^4+5x^3+4x^2+6x-4=left(x^2+frac{5}{2}xright)^2-left(frac{9}{4}x^2-6x+4right)=$$
                  $$=left(x^2+frac{5}{2}xright)^2-left(frac{3}{2}x-2right)^2=(x^2+x+2)(x^2+4x-2).$$
                  Can you end it now?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 31 at 22:46









                  Michael RozenbergMichael Rozenberg

                  108k1895200




                  108k1895200






























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