hrcheerlove
Multiply the polynomials:
1. (a + 3)(a - 2)
2. (x2 + 4)(x2 - 4)
3. (x2 + 3x + 1)(x2 + x + 2)
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Outkast3r09
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use the distributive property
\[a(b\pm c)=ab\pm ac\]
Outkast3r09
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using this you take the first variable of the first and multiply it by the second parenthesis and same with the second variable/constant of the first
\[a(a-2)+3(a-2)\]
Outkast3r09
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what do you get when yo udo this?
hrcheerlove
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So then i got: 2a-2a + 3a-6
hrcheerlove
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But I dont think that is right lol
hrcheerlove
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@Outkast3r09
Outkast3r09
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everything but your first multiplication is correct
\[a(a)\neq 2a\]
Outkast3r09
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hint: when you have a number to a power you get this correct
\[2^2=2*2\]
Outkast3r09
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so \[a*a=?\]
hrcheerlove
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uhhh 2 squared ???
hrcheerlove
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@Outkast3r09
Outkast3r09
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not quite so 2 is the base
Outkast3r09
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in \[2^2\]
Outkast3r09
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the power tells you how many times you multiply that base by itself
Outkast3r09
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so \[2^3=2*2*2\]
\[2^5=2*2*2*2*2\]
Outkast3r09
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so in your case \[
a*a\]
a is your base and the number of bases you multiply is your power
Outkast3r09
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so \[a*a=a^2\]
hrcheerlove
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I know how to do powers lol. I just dont know what \[a \times a\] is
hrcheerlove
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oh okay lol
some_someone
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|dw:1359943457681:dw|
Outkast3r09
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I think it's just the variable thats giving you a tough time if it helps let a be whatever you want and then replace it back with a.
For example i have
\[a*a*a*a\]
in your head think a=5
\[5*5*5*5=5^4\]
replace 5 with a
some_someone
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|dw:1359943594086:dw|
Outkast3r09
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some someone let her do the last one
hrcheerlove
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so would it be \[a ^{2}-5a -6\] ?
Outkast3r09
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\[-2a+3a\]
if you have 3 dollars and lost 2 dollars, how many do you have now =]
hrcheerlove
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lol 1 dollar
some_someone
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|dw:1359943710986:dw|
some_someone
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|dw:1359943887804:dw|
hrcheerlove
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Thank you thank you thank you @Outkast3r09 and @some_someone !!!!
some_someone
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your welcome hope we could help you :)