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OregonDuck
 one year ago
MEDAL AND FAN
GROUPING BY FACTORING
2x^2+4+1+4
OregonDuck
 one year ago
MEDAL AND FAN GROUPING BY FACTORING 2x^2+4+1+4

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Nnesha
 one year ago
Best ResponseYou've already chosen the best response.02x^2 +4x+1x +4 like this ?

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.02x^2 +4x+1x +4 yes like that

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Combine 4x and x to make 5x, then you will have 2x^2+5x+4

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0no factoring by grouping

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0nice make a group of two terms dw:1437747748496:dw what is common in red box ?

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Oh okay, find the common term in each binomial

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1As nnesha said, the first box would have 2x, the 2nd box would have 1

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0cute and ? what about variable

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0x^2 is same as x times x so x is also common

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1@Oregon the first box has a common term of 2x, there is an x^2 and x in first box

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437747924145:dw take out common factor 2x what will go in the parentheses ?

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1So you would have \[2X(x+2) + 1(x+4)\]

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1To get the parathenses inside, you have to divide the binomial by 2x

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Using that you have to group the remainders (2x and 1) to get (2x+1)(x+2)(x+4)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0i'll let him help good luck

OregonDuck
 one year ago
Best ResponseYou've already chosen the best response.0so what is the answer?
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