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Mathematics 17 Online
OpenStudy (anonymous):

Solve the following equation: 8y^3 − 2y^2 − 6y = 0

OpenStudy (anonymous):

@hello1213

OpenStudy (anonymous):

what is common between all of the terms?

OpenStudy (anonymous):

2y?

OpenStudy (anonymous):

yes! so now we have\[2y(4y^2-2y-6)\]

OpenStudy (anonymous):

yup, =0, now what?

OpenStudy (anonymous):

\[=2y \left(4 y^2-y-3 \right)\]

OpenStudy (anonymous):

yeah the middle term should be y not 2y.

OpenStudy (anonymous):

I got it, now I factor it out more?

OpenStudy (anonymous):

4*-3=-12 4-3=1 4*-3=-12

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

4-3=1 is the right one

OpenStudy (anonymous):

2y(y+4)(y-3)?

OpenStudy (anonymous):

2y+4?

OpenStudy (anonymous):

2 y (y-1) (4 y+3)=0 y= 0, 1 and -3/4

OpenStudy (anonymous):

thank you ! :)

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