Multiplying vs. Squaring Binomials

[mstcrow5429]

Why do you have to handle (x + 3)^2, or (x + 3)(x + 3) differently than (x + 3)(x + 4)?

[Blue]

what do you mean by "differently"?

[mstcrow5429]

Well, non-squared would be (A^2 + B^2) or (A^2 - B^2), whereas squared, it's (A + B)2 = A2 + 2AB + B2; (A - B)2 = A2 - 2AB + B2

[Blue]

Well, non-squared would be (A^2 + B^2) or (A^2 - B^2), whereas squared, it's (A + B)2 = A2 + 2AB + B2; (A - B)2 = A2 - 2AB + B2

i hope i understand your question...if not, just ignore my explanation

so (x + 3)^2 or (x + 3)(x + 3) is different from (x +3)(x +4) b/c 3 and 4 are different numbers. in order for the (A + B)^2 rule to work, you have to think of A and B as variables, in which each variable represents ONE number.

so if you compare (A + B)^2 to (x + 3)^2, A = x and B = 3
but in the (x + 3)(x + 4) problem, A still equals "x" but you can't have B equaling both 3 and 4. it would be more like (A + B)(A + C). you know what i mean?...one number per variable.

you can minimize (x + 3)(x + 3) to (x + 3)^2 b/c what's in each parentheses is the exact same thing but you CAN'T minimize (x + 3)(x + 4) b/c each parentheses is different...
thus your (A + B)2 = A2 + 2AB + B2 doesn't apply to (x + 3)(x + 4).


another way to look at it by applying the (A + B)2 = A2 + 2AB + B2 rule:

(x + 3)^2 comes out to be x^2 + 6x + 9 right?
do some substitution:
A = x
B = 3
isn't x^2 + 6x + 9 the same as saying x^2 + 2(x)(3) + 3^2?

but you can't do that for (x + 3)(x + 4) b/c look at this:
answer: x^2 + 7x + 12
A = x but you can't have B equaling 3 and 4 because no matter how you multiply 3 or 4 by 2 or even squaring the numbers, they won't come out to 7 or 12 respectively right?
thus the rule does NOT work for (x + 3)(x + 4)


sorry for the lengthy explanation. i hope it addresses your question. if not, sorry about that! :p