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Mathematics 9 Online
OpenStudy (anonymous):

factor 6y^6-5y^3-4

OpenStudy (anonymous):

this might help to y^3=x 6x^2-5x-4 then factor this and sub back in y

OpenStudy (anonymous):

Sorry I wasn't specific but factor in Quadratic Form

OpenStudy (anonymous):

idk how to do quadratic form

OpenStudy (anonymous):

This is a commonly used method; you know that it must add to be -5, and multiply to be -24. So you get \[6y ^{6} + 3y ^{3} -8y^{3}-4\] From there, you can easily factor out, so your final solution should be \[(3y ^{3}-4)(2y ^{3} + 1)\]

OpenStudy (campbell_st):

make it a quadratic by substituting u = y^3 then you have 6u^2 - 5u - 4 = (3u-4)(2u + 1) replace the u with y^3 (3y^3 -4)(2y^3 +1)

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