Question 1048380
<pre><b>This must be an identity

{{{x^4+ax^3-3x^2+2ax+a^2}}}{{{""=""}}}{{{((x+2)(x+k)^"")^2}}}

If x=0

{{{(0)^4+a(0)^3-3(0)^2+2a(0)+a^2}}}{{{""=""}}}{{{(((0)^""+2)((0)^""+k)^"")^2}}}

{{{a^2}}}{{{""=""}}}{{{(2k)^2}}}

{{{a}}}{{{""=""}}}{{{"" +- 2k}}}

If x=1

{{{(1)^4+a(1)^3-3(1)^2+2a(1)+a^2}}}{{{""=""}}}{{{(((1)^""+2)((1)^""+k)^"")^2}}}

{{{1+a-3+2a+a^2}}}{{{""=""}}}{{{((1+2)(1+k)^"")^2}}}

{{{1+a-3+2a+a^2}}}{{{""=""}}}{{{(3(1+k)^"")^2}}}

{{{1+a-3+2a+a^2}}}{{{""=""}}}{{{9(1+k)^2}}}

Substituting {{{a}}}{{{""=""}}}{{{2k}}}

{{{1+(2k)-3+2(2k)+(2k)^2}}}{{{""=""}}}{{{9(1+k)^2}}}

{{{4k^2+6k-2}}}{{{""=""}}}{{{9k^2+18k+9}}}

{{{-5k^2-12k-11}}}{{{""=""}}}{{{0}}}

{{{5k^2+12k+11}}}{{{""=""}}}{{{0}}}

Discriminant = {{{12^2-4*5*11=144-220=-76}}}

So it has imaginary solutions.  We ignore {{{a}}}{{{""=""}}}{{{2k}}}


Substituting {{{a}}}{{{""=""}}}{{{-2k}}}

{{{1+(-2k)-3+2(-2k)+(-2k)^2}}}{{{""=""}}}{{{9(1+k)^2}}}

{{{1-2k-3-4k+4k^2}}}{{{""=""}}}{{{9(1+2k+k^2)}}}

{{{4k^2-6k-2}}}{{{""=""}}}{{{9k^2+18k+9}}}

{{{-5k^2-24k-11}}}{{{""=""}}}{{{0}}}

{{{5k^2+24k+11}}}{{{""=""}}}{{{0}}}

Discriminant = {{{24^2-4*5*11=576-220=356}}}

That has real solutions

{{{5k^2+24k+11}}}{{{""=""}}}{{{0}}}

{{{k}}}{{{""=""}}}{{{(-24 +- sqrt(24^2-4*5*11 ))/(2*5) }}}

{{{k}}}{{{""=""}}}{{{(-24 +- sqrt(356 ))/10 }}}

{{{k}}}{{{""=""}}}{{{(-24 +- sqrt(4*89))/10 }}}

{{{k}}}{{{""=""}}}{{{(-24 +- 2sqrt(89))/10 }}}

{{{k}}}{{{""=""}}}{{{(2(-12 +- sqrt(89)))/10 }}}

{{{k}}}{{{""=""}}}{{{(-12 +- sqrt(89))/5 }}} 

{{{a}}}{{{""=""}}}{{{-2k}}}

{{{k}}}{{{""=""}}}{{{(-12 +- sqrt(89))/5 }}}

{{{a}}}{{{""=""}}}{{{-2*expr((-12 +- sqrt(89))/5) }}}

{{{a}}}{{{""=""}}}{{{(24 +- 2sqrt(89))/5 }}}

So

{{{((x+2)(x+k)^"")^2}}}

{{{((x+2)(x+((-12 +- sqrt(89))/5))^"")^2}}}

{{{expr(1/25)((x+2)^""(5x-sqrt(89)-12))^2}}}

Edwin</pre></b>