Question 1025959
<pre> 
{{{x^2- (((a+b)^2+a^2)/(a^2+ab))x+1}}}

{{{x^2- (((a+b)^2+a^2)/(a(a+b)))x+1}}} 

Since a+b appears twice, let c = a+b 

{{{x^2- ((c^2+a^2)/(ac))x+1}}}

Get a least common denominator of ac

{{{expr((ac)/(ac))x^2- ((c^2+a^2)/(ac))x+1*expr((ac)/(ac))}}}

Factor out {{{1/ac}}}

{{{expr(1/ac)((ac)x^2-(a^2+c^2)^""x+ac)}}}

So we try to factorize the trinomial like this,
breaking up {{{(ac)x^2}}} as the product of {{{ax}}} and {{{cx}}}:

{{{expr(1/ac)



(matrix(1,4,ax^"",""-"","",""))(matrix(1,4,cx,""-"","","")^"" 

)}}}

It doesn't take too much thinking to know that we
must fill in the blank spaces with c and a to
cause the product to be ac

{{{expr(1/ac)



(matrix(1,3,ax^"",""-"",c))(matrix(1,3,cx^"",""-"",a) 

)}}}

And if we find the outer and inner products, we
find that we get -aČ-bČ which is -(aČ+bČ).

So that is the correct factorization.  All that's
left is to substitute (a+b) for c and simplify:

{{{expr(1/(a(a+b)))



(matrix(1,3,ax^"",""-"",(a+b)))(matrix(1,3,(a+b)x^"",""-"",a) 

)}}}  

Removing the inner parentheses we have:

{{{expr(1/(a(a+b)))



(matrix(1,5,ax^"",""-"",a,""-"",b))(matrix(1,5,ax,""+"",bx^"",""-"",a))}}} 

Edwin</pre>