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The product of (a − b)(a − b) is a^2 − b^2. A: Sometimes B: Always C: Never *is it a?

Mathematics
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I think it is B because it would end up being a^2-ab-ba+b^2 Unless there is a zero of course so then it could be A
The product of (a-b)(a-b) is a^2-2ab+b^2 now what conclusion do you think.
so its never hence C

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Other answers:

it cant be b because i choose it and i got it wrong, but i stuck between a and c
Yes C sorry I got the letter mixed up
okay well thank you and can you help me with 1 more?
The answer is never
Find the two products below. Compare and contrast, in complete sentences, the similarities and differences of the two. (x + 4)(x − 4) and (x + 4)(x + 4)
the first would play out to x^2-4x+4x-16 leaving you with x^2-16 The second would become x^2+4x+4x+14 leaving you with x^2+8x+14 See the difference?
So the second is +16 not +14
Do you understand how to distribute those types of problems?
no im confused?
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the first x multiplies to each term in the second set of parenthesis. Then the first 4 multiplies to each term in the second set of parenthesis. Add all the terms together and you are good to go :)
ohh okay! i get it
(x + 4)(x + 4) x^2 + 4 + 4 + 16 this is what i get for the second equation
Yes but remember to carry x along. The second and third terms you have just 4. But the 4 multiplies with x making them each 4x So x^2+4x+4x+16
You will be expected to combine like terms as well. So go ahead and add the 4xs together giving you 8x making your final result for that problem x^2+8x+16
oh okay can you help me out with the similarities and differences of the two equations?
You can see from our calculations that it is the minus sign that causes the major differences. It turns positive 16 into negative 16 and instead of creating 8x with the 4xs, the 4xs end up canceling each other out because 1-1=0 so 4x-4x=0 as well.
Notice though that both problems do yield an x^2 and a 16(just a negative 16 for one of them)
oh yea thats true well thank you soo much, you have been a great help!(:
No problem :) good luck!
Thanks(:

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