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Mathematics 18 Online
OpenStudy (anonymous):

Factor by grouping equation is (4g2 + 12gh + 5g + 15h)?

OpenStudy (paxpolaris):

take the first two terms and remove the common out ... then the last two

OpenStudy (anonymous):

(4g2+12gh)+(5g+15h) 4g(gh+3h)+5(g+3h) not sure this is correct but its as far as i can go

OpenStudy (anonymous):

is simple

OpenStudy (anonymous):

i have a extra h in the first part sorry

OpenStudy (paxpolaris):

\[\Large 4g(g+3h)+5(g+3h)\]

OpenStudy (paxpolaris):

then remove the common (g+3h) out

OpenStudy (anonymous):

so its (g + 3h)(4g + 5)? thats the answer right?

OpenStudy (anonymous):

would this turn to (4g+5) (g+3h)^2

OpenStudy (paxpolaris):

yes :) @alamojoe9

OpenStudy (paxpolaris):

no ^2

OpenStudy (anonymous):

(4g-5h)(g+3) ya that is correct

OpenStudy (anonymous):

thanks!

OpenStudy (anonymous):

but dont you have to account for both (g+3h)

OpenStudy (anonymous):

the first one poster was the right one which is "so its (g + 3h)(4g + 5)"

OpenStudy (paxpolaris):

@Nismo450Z let, \(\large(g+3h) = P\) \[a \cdot P +b \cdot P =(a+b)P\]

OpenStudy (anonymous):

now i see i will keep that in mind thanks

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