Try multiplying everything out, then combining:
(a+1)^2 - (b-3)^2 =
(a+1)(a+1) - (b-3)(b-3) =
a^2 + 2a + 1 - (b^2 - 6b + 9) =
a^2 + 2a + 1 - b^2 + 6b - 9 =
a^2 + 2a - b^2 + 6b - 8
Not sure you can go much further with it than that.
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EDITED TO ADD: Answerers #3 (Steiner) and #6 (P.Rutherford) have it! Answerer #3 gave the hint, Answerer #6 did the work. If you want full credit on this homework problem, go with Answerer #6.
2006-10-12 14:38:14
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answer #1
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answered by I ♥ AUG 6
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Okay (a+1)^2 - (b-3)^2 factored:
Let x = (a+1) and y = (b-3), so we have:
x^2^- y^2 factoring this is simple:
(x+y)(x-y) now get the original variables back:
[(a+1) + (b-3)][(a+1) - (b-3)] simplify
[a + b -2][a - b +4] Voila!
2006-10-12 21:44:15
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answer #2
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answered by p_rutherford2003 5
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first, since both the terms are squares you can find the square root of the problem. (hint: the square root of a number squared is the number itself) so then you should have (a+1) - (b-3). then since in both parenthesis is just addition and subtraction you can ignore them so you would get a-b+4, because positive 1 minus negitive 3 = 4
hope this helped
2006-10-12 23:07:07
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answer #3
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answered by princess 2
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Hint: You have 2 squares. How do you factor their difference?
2006-10-12 21:37:55
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answer #4
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answered by steiner1745 7
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This might help, use FOIL.
First, Outside, Inside, Last.
(a+1)(a+1)
(a*a)+(a*1)+(1*a)+(1+1)
F O I L
Hope that helps.
2006-10-12 21:37:01
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answer #5
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answered by frozenpancake582 4
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I am not competley sure but I have an idea.
(a + 1)^2 - (b - 3)^2
change signs
a + 1^2 + -b + -3^2
a+1 + -b + 9
a+10 +-b
2006-10-12 21:40:38
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answer #6
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answered by ^..^ 4
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If you mean (a+1)^2 - (b-3)^2 then:
(a+1)(a+1)=a^2 + 2ab + b^2
(b-3)(b-3)=b^2 - 6b +9
(a^2 + 2ab +b^2) - (b^2 - 6b +9)
= a^2 + 2ab +b^2 - b^2 + 6b -9
=a^2 + 2ab +6b -9
2006-10-12 21:46:33
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answer #7
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answered by fathotra 1
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srry. I'm not good at math.
2006-10-12 21:36:13
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answer #8
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answered by ruty 1
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