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Multiplying vs. Squaring Binomials - Printable Version +- Online Degrees and CLEP and DSST Exam Prep Discussion (https://degreeforum.net/mybb) +-- Forum: Miscellaneous (https://degreeforum.net/mybb/Forum-Miscellaneous) +--- Forum: Off Topic (https://degreeforum.net/mybb/Forum-Off-Topic) +---- Forum: What does this Flashcard mean or Do this math problem for me (https://degreeforum.net/mybb/Forum-What-does-this-Flashcard-mean-or-Do-this-math-problem-for-me) +---- Thread: Multiplying vs. Squaring Binomials (/Thread-Multiplying-vs-Squaring-Binomials--2229) |
Multiplying vs. Squaring Binomials - mstcrow5429 - 08-14-2007 Why do you have to handle (x + 3)^2, or (x + 3)(x + 3) differently than (x + 3)(x + 4)? Multiplying vs. Squaring Binomials - Blue - 08-14-2007 what do you mean by "differently"? Multiplying vs. Squaring Binomials - mstcrow5429 - 08-14-2007 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 Multiplying vs. Squaring Binomials - Blue - 08-14-2007 mstcrow5429 Wrote: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 |