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OpenStudy (anonymous):
Solve: x(10x+3)(x+4)=0
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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 ?
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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
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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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