Hey (: I need help figguring out the process for an equation!! please help?? 5. Factor completely 21x3 + 35x2 + 9x + 15. (3x − 5)(7x2 − 3) (3x − 5)(7x2 + 3) (3x + 5)(7x2 − 3) (3x + 5)(7x2 + 3)
factor by grouping
\(\bf 21x^3 + 35x^2 + 9x + 15\\ (21x^3 + 35x^2)+(9x + 15)\\ 7x^2(3x+5)+3(3x+5)\impliedby \textit{take common factor of it}\)
okay!! not to certain how to properly do that!! i know the answer and i have the process. i just need to know how it all works!! how do you add a picture to this??
\(\bf 21x^3 + 35x^2 + 9x + 15\\ (21x^3 + 35x^2)+(9x + 15)\\ 7x^2{\color{blue}{ (3x+5)}}+3{\color{blue}{ (3x+5) }}\impliedby \textit{see any common factors?}\)
yeah!!! how do you start with the equation what was your first step??
as acxbox22 said, by grouping then take any common factors from the groups
you have 4 terms 1 2 3 4 then we grabbed (1 2) (3 4) then common factors of them
yeap, that's correct
okay!! (: did you get the picture?? y
\(\bf 21x^3 + 35x^2 + 9x + 15\\ (21x^3 + 35x^2)+(9x + 15)\\ 7x^2{\color{blue}{ (3x+5)}}+3{\color{blue}{ (3x+5) }} \\ \quad \\ (7x^2+3)(3x+5)\)
yes
i see you got the same thing as she did!!! so lets see you took the numbers and put them into perethesis....
then you took the numbers and found the greatest common multipul??
yeap
okay after that then what did you do??
\(\bf 21x^3 + 35x^2 + 9x + 15 \\ \quad \\ (21x^3 + 35x^2)+(9x + 15)\impliedby grouping \\ \quad \\ 7x^2{\color{blue}{ (3x+5)}}+3{\color{blue}{ (3x+5) }}\impliedby \textit{common factor on each group} \\ \quad \\ (7x^2+3)(3x+5)\impliedby \textit{common factor on the terms}\)
once you take the common factor on each group what's left inside from the 1st group, is "common" with what's left inside the 2nd group then you take common factor of that
ah!!! I think i get it now!!!(: thanks so so much to both of you. is there a way to give both of you medals??XD
yw
yw??
that's okay!! Just thanks so so much!!! and see you both later!!! (:
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