multiply[(3a-(2a-3)][3a+(2a-3)] how do you work this problem when it has the ext
ID: 3089250 • Letter: M
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multiply[(3a-(2a-3)][3a+(2a-3)] how do you work this problem when it has the extra minus andplus signs? 3a - (2a-3) and 3a + (2a - 3). Is it the same procedure? I got [6a-9a][6a-9a] but Idon't think it is correct. Please help me understand this stuff : { multiply[(3a-(2a-3)][3a+(2a-3)] how do you work this problem when it has the extra minus andplus signs? 3a - (2a-3) and 3a + (2a - 3). Is it the same procedure? I got [6a-9a][6a-9a] but Idon't think it is correct. Please help me understand this stuff : { how do you work this problem when it has the extra minus andplus signs? 3a - (2a-3) and 3a + (2a - 3). Is it the same procedure? I got [6a-9a][6a-9a] but Idon't think it is correct. Please help me understand this stuff : {Explanation / Answer
[3a-(2a-3)] it is the same as having a one in front of(2a-3). So it wuold look like this [3a-1(2a-3)] so you woulddistribute the -1 to be (3a-2a+3) = (a+3). [3a+(2a-3)] for this there is really nothing to do since it isthe same thing as having a +1 in front of the parenthesis :[3a+1(2a-3)] = (3a+2a-3) = (5a-3) So now you will have (a+3)(5a-3) =(a)(5a)+(a)(-3)+(3)(5a)+(3)(-3) = 5a2-3a+15a-9 =5a2+12a-9 Feel free to email me with questions and please rate.Related Questions
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