if you are using algebra tiles to create trinomial rectangles and perfect trinomial squares,and you take a trinomial like 2x^2+8x+8 and factor it,you get 2(x+2) (x+2) which is a perfect square,yet when you try to construct the area of this trinomial,you can not make a perfect square out of it,also when i construct other squares such as ones that have square roots like (2x+2)^2.or (3x+3)^2,they expand into perfect squares yet,when you factor them you get binomial factors like 4(x+1)(x+1) or 9(x+1)(x+1) are these just factorization bugs?,can 2x^2+8x+8 be constructed into a perfect square?, 2x^2 is not the square of x,because 2 is not a square number and has no rational square root,does anybody want to clear this up?.......thank you hey stapel eliz,i re-edited this ,because i wrote down 2(x+1)(x+1) and 3(x+1) instead of 4(x+1) and 9(x+1) this type of factorization is somewhat unfamiliar

because iam pulling out a GCF first,before i split the middle term and go into factoring by grouping,this stand alone number outside of the binomial factors is somewhat weird looking

although i know they are square numbers,with the exception of course of 2(x+1)(x+1),you are right though 2(x+1)(x+1) is not a square,i guess then my real question is

does a number hanging outside of the binomial factors represent a factor member of the binomial pairs or is it its own non square idenity,i think you answered that question

when you said (x+2)^2 is a perfect square but 2(x+2)^2 is not although 2^2(x+2)^2 is a perfect square,when you start these tile toys out,you can start them out using

binomial factors such as (2x+2)^2 or (3x+3)^2 or (4x+4)^2,yet you can never re factor these trinomial products back into these original binomial factors without a number

hanging outside the binomial factors,has anything ever been taught on "a protocol" to repackage these binomial factors back into there original roots' without the multiplying

number hanging outside of the binomial perenthesis?,i find this number annoying as it destroys the simplicity of the binomial factors themselves,or is this something you

are left to figure out on your own?