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08-14-2007, 07:17 PM
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Multiplying vs. Squaring Binomials
Why do you have to handle (x + 3)^2, or (x + 3)(x + 3) differently than (x + 3)(x + 4)?
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Excelsior, BS Finance, pursuing degree Studying: DSST P. of Stat. Completed: CLEP: Hum. (67), Hist. of U.S. I (74), Hist. of U.S. II (71), Intro. Psych. (69), Intro. Soc. (72), Soc. Sci. and Hist. (74), Western Civ I (72), Western Civ II (70), Am. Lit. (60), Intro. to Educ. Psych. (62), P. of Management (74), P. of Market. (74), Intro. Bus. Law (67), P. of Accounting (60), AmGov (68) DSST: Ethics in Am. (76), P. of Super.(67), HRM (65), Intro to Bus. (70), MIS (65), P. of Fin (62), M&B (65) ECE: OB (B) Total Credits: 105     
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08-14-2007, 07:49 PM
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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
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Excelsior, BS Finance, pursuing degree Studying: DSST P. of Stat. Completed: CLEP: Hum. (67), Hist. of U.S. I (74), Hist. of U.S. II (71), Intro. Psych. (69), Intro. Soc. (72), Soc. Sci. and Hist. (74), Western Civ I (72), Western Civ II (70), Am. Lit. (60), Intro. to Educ. Psych. (62), P. of Management (74), P. of Market. (74), Intro. Bus. Law (67), P. of Accounting (60), AmGov (68) DSST: Ethics in Am. (76), P. of Super.(67), HRM (65), Intro to Bus. (70), MIS (65), P. of Fin (62), M&B (65) ECE: OB (B) Total Credits: 105
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08-14-2007, 11:01 PM
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Quote:
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! 
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Jenny =]
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