factor the expression c^2+4c+4
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basically you have to find 2 numbers that a) multiply to 4 AND b) add to 4
those two numbers are 2 and 2 since 2+2 = 4, 2*2 = 4 so that's why c^2 + 4c + 4 factors to (c+2)(c+2)
A. (c + 2)^2 B. (c + 1)(c + 4) C. (c^2 + 4)^2 D. the expression cannot be factored because it is prime.
I am confused can you break it down for me again please
(c+2)(c+2) is the same as (c+2)^2
since something like x^2 means x times x
so would it be a? (c+2)^2 which is c^2+4
(c+2)^2 is NOT c^2 + 4
but yes, it is choice A
try using the quadratic formula. maybe she knows how to use that. you will get one root c=-2. therefore you know c=-2 has multiplicity of 2 therefore we factor it as (c+2)^2
oh ok thank you, but how are they not the same if when you solve (c^2+2)^2 it makes 2x2+C times C and then simplify = 4+c^2
(c^2+2)^2 is not the same as (c+2)^2
expanding (a+b)^2=a^2+2ab+b^2 in general terms
oh
so if you expand (c+2)^2 you get c^2+2(c)(2)+2^2 =c^2+4c+4
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