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lgbasalloteBest ResponseYou've already chosen the best response.1
2x^2  3x  9 break down the middle term 2x^2 + 6x  3x  9 group them (2x^2 + 6x)  (3x + 9) factor out 2x(x + 3)  3(x+3) do you see it now?
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.1
\[x^23x9=0\] \[x^26x+3x9=0\] \[2x(x3)+3(x3)=0\] now you try to simplify.
 one year ago

annasBest ResponseYou've already chosen the best response.0
you can use the quadratic formula i.e \[b \pm \sqrt{b ^{2}4ac}/2a\] here a=2 b=3 c=9 put the values of a, b and c in equation and you'll get your answer
 one year ago

joemarieBest ResponseYou've already chosen the best response.2
thanks for helping me
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.1
we have to make the first term of these sums.
 one year ago

joemarieBest ResponseYou've already chosen the best response.2
hi what is the factor of 8x^3216
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.1
\[First Term= (Middle Term)^2/4*(Last Term)\]
 one year ago

joemarieBest ResponseYou've already chosen the best response.2
thanks,but im not asking what is the procedure im asking is what is the solution?
 one year ago

joemarieBest ResponseYou've already chosen the best response.2
oo nga pala tuloy ung sa e nglish
 one year ago

muhammad9t5Best ResponseYou've already chosen the best response.1
factor of 8x^3216 \[8x(x^227)=0\] \[8x=0 \] \[x=0\] or \[x^227=0\] \[x^2=27\] \[x=\sqrt{27}\]
 one year ago

joemarieBest ResponseYou've already chosen the best response.2
abano add mo ko sa fwend m!!
 one year ago

WiredBest ResponseYou've already chosen the best response.0
@danny_dumond Please don't post only answers as that's against the Code of Conduct: http://openstudy.com/codeofconduct
 one year ago

timo86mBest ResponseYou've already chosen the best response.0
well find something like (cx+a) (dx+b) to = 2x^23x9
 one year ago
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