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please help me!!

quadratic equations:
there are 4 method to factor out the quadratic equations

s*

for the first one I would start it as
(2a+ ?b) (a+?b)

so i would multiply the 2 and the 3 to get the middle term?

that would be 2a^2+8ab+6b^2 thats not right

hmm

okay im not following why do i add?

or would this one be prime?

1 and 6

\[\huge\rm \color{Red}{2a^2+6ab}+\color{blue}{1ab+3b^2}\]
what is common in first two terms ?

so that makes (a+3b) (2a+b) right?

because you factor out the highest factor

of each group

yes that's correct!

okay cool

you can foil it to check your answer
(2a+b)(a+3b)=2a^2+7ab+3b^2

do you want to try it first ? :)

im trying the other ab one first.

okay :=)

okay

so 6 times 14 = 84 not negative
but what if take -14 times 6 =

that equals the -8 that is needed

\[\rm 3a^2+6ab-14ab+28b^2\]

okay like this 3a^2 -14ab+6ab-28b^2

then i separate them into groups and factor

yes right! :+)

yes right

okay now on to the last problem this one i have started it as |dw:1440355529657:dw|

wait hold on i think i just figured it out

how do you know it would be one minus and one plus parentheses ?
i mean \[\huge\rm (6x-?)(x+??)\]

well thank you natasha

|dw:1440356037300:dw|
that's right !

|dw:1440356266572:dw|