Question 1028739
<pre>
We learn that 

{{{A^2+2AB+B^2}}} 

factors as 

{{{(A+B)^2}}}. 

Therefore when we are trying to factor a trinomial and we 
observe that the first and third terms of the trinomial 
happen to be perfect squares:

We always then check to see if the middle term happens to be 
twice the product of their square roots.  For if so, then the
trinomial factors as the square of a binomial.

We are trying to factor the trinomial:

{{{matrix(2,5,"","","","","",

x^(4/3),

""+"", 

1/2, 

""+"", 

expr(1/16)x^(-4/3)  ) }}}

The first term is the square of {{{matrix(2,1,"",x^(2/3))}}}
The third term is the square of {{{matrix(2,1,"",expr(1/4)x^(-2/3))}}}, so we should treat 
this trinomial just as we would treat any trinomial whose
first and last terms are perfect squares.

We find twice the product of their square roots:

{{{matrix(2,1,"",2*(x^(2/3))*(expr(1/4)x^(-2/3)))}}}

The exponents add to zero and {{{x^0}}} is 1, so above we see that twice
the product of their square roots is {{{1/2}}}, which is the middle term 
of the trinomial.  So the factorization of the trinomial is

{{{(matrix(2,3,"","","",
x^(2/3), ""+"", expr(1/4)x^(-2/3)))^2 }}}

Edwin</pre>