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Mathematics 8 Online
OpenStudy (anonymous):

Factor completely

OpenStudy (anonymous):

\[m^3n-8m^2n^2+12mn^3\]

jhonyy9 (jhonyy9):

there is m2n and m2n2 and mn2 ?

OpenStudy (anonymous):

the 1st is m^3n and then 2nd one is correct and 3rd is mn^3

jhonyy9 (jhonyy9):

ok

jhonyy9 (jhonyy9):

m3n -8m2n2 +12mn3 yes ?

OpenStudy (anonymous):

yes that is correct.

jhonyy9 (jhonyy9):

mn(m2 -8mn +12n2)

jhonyy9 (jhonyy9):

so than m2-8mn +12n2 =(m-2n)(m-6n)

jhonyy9 (jhonyy9):

than this all will be mn(m-2n)(m-6n)

jhonyy9 (jhonyy9):

ok ?

jhonyy9 (jhonyy9):

do you understand it ?

OpenStudy (anonymous):

sorta but not really sorry im terrible at math :/

jhonyy9 (jhonyy9):

what is not clearly ?

OpenStudy (anonymous):

what is the first step

jhonyy9 (jhonyy9):

first you see that mn is the common term in every terms so than this can be extracting than you get mn(....)

jhonyy9 (jhonyy9):

ok ?

jhonyy9 (jhonyy9):

and after this inside parantheses you have m2-8mn+12n2 so here you need get two numbers a and b so than a+b =-8 and a*b=12

jhonyy9 (jhonyy9):

a+b=-8 so -6-2=-8 and -6*-2=12

jhonyy9 (jhonyy9):

right ?

jhonyy9 (jhonyy9):

now is clearly ?

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