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Mathematics 21 Online
OpenStudy (anonymous):

LCM of x^2+x-2 and x^2+8x+12

OpenStudy (anonymous):

first one factors as \[x^2+x-2=(x+2)(x-1)\] second one as \[x^2+8x+12=(x+2)(x+6)\] so LCM is \[(x-1)(x+2)(x+6)\]

OpenStudy (anonymous):

factor each of the following 1.) (x+2)(x-1) 2.) (x+6)(x+2) The common ones in each count once so (x-2) is a factor then the different ones in each factor count once so (x-1) and (x+6). So the LCM is (x+2)(x-1)(x+6)

OpenStudy (anonymous):

(x+2) sorry not (x-2).

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