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Mathematics 20 Online
OpenStudy (anonymous):

Solve: x(10x+3)(x+4)=0

OpenStudy (anonymous):

@dpasingh

OpenStudy (a1234):

x(10x+3)(x+4)=0 x(10x^2 + 40x + 3x + 12) = 0 10x^3 + 40x^2 + 3x^2 + 12x = 0 10x^3 + 43x^2 + 12x = 0 Does that look right?

OpenStudy (anonymous):

@hoblos does that look right ^ ?

OpenStudy (hoblos):

what you did is expanding not solving to solve means to find the value of x so that x(10x+3)(x+4) is equal to 0

OpenStudy (anonymous):

so how can I do that ?

OpenStudy (hoblos):

set each factor =0 and solve x=0 10x+3=0 x+4=0 solve these three equations

OpenStudy (a1234):

Yeah, sorry, I did not exactly know how to do that. That's why I asked.

OpenStudy (anonymous):

Alright Lol Im kind of rushing I have two more questions and I leave to class in 10 min

OpenStudy (hoblos):

you got it @cancherolugano ?

OpenStudy (anonymous):

yea but I just don't know how to show the work for it

OpenStudy (hoblos):

x=0 10x+3=0 so 10x=-3 , x=-3/10 x+4=0 so x=-4 thus the solution set is {-4, -3/10, 0}

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