Question 1077643
<pre>
If {{{x^4+2x^3+6x^2+5x+6}}} is identical to (x^2+x)^2+a(x^2+x)+b,
then any value substituted for x in both will always give the 
same answer.

Substituting x=0 in the first gives

{{{0^4+2*0^3+6*0^2+5*0+6}}}{{{""=""}}}{{{6}}}

Substituting x=0 in the second gives

{{{(0^2+0)^2+a(0^2+0)+b}}}{{{""=""}}}{{{b}}}

So b=6

Substituting x=1 in the first gives

{{{1^4+2*1^3+6*1^2+5*1+6}}}{{{""=""}}}{{{1+2+6+5+6}}}{{{""=""}}}{{{20}}}

Substituting x=1 in the second gives

{{{(1^2+1)^2+a(1^2+1)+b}}}{{{""=""}}}{{{(2)^2+a(2)+b}}}{{{""=""}}}{{{4+2a+b}}}

So 4+2a+b = 20, and since b=6
   4+2a+6 = 20
    2a+10 = 20
       2a = 10
        a = 5

So

{{{x^4+2x^3+6x^2+5x+6}}} is identical to {{{(x^2+x)^2+a(x^2+x)+b}}} 

and we substitute a=5 and b=6:

{{{(x^2+x)^2+5(x^2+x)+6}}}.

And that's the form desired.

To factor it let u = x²+x

{{{u^2+5u+6}}}

That factors as

{{{(u+3)(u+2)}}}

Substitute x²+x for u

{{{(x^2+x+3)(x^2+x+2)}}}

Edwin</pre>