anonymous
  • anonymous
2y(y^2-3y-2)-3(5y+2y^2-y^3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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sandra
  • sandra
ok so you need to distribute out each term
sandra
  • sandra
so 1. 2y(y^2-3y-2) = (2y*y^2) - (2y *3y) - (2y*2) = 2y^3 -6y^2 -4y
amistre64
  • amistre64
id "factor" out each term; but feel free to distribute them out :)

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sandra
  • sandra
hmmm, good point
anonymous
  • anonymous
thank u!
sandra
  • sandra
well I'll do it this way, then you sow that way =)
sandra
  • sandra
2. 3(5y+2y^2-y^3) = (3*5y) + (3*2y^2) - (3*y^3) = 15y + 6y^2 - 3y^3
amistre64
  • amistre64
youre doing fine :)
sandra
  • sandra
now subtract them: 2y^3 -6y^2 -4y - (15y + 6y^2 - 3y^3) = 2y^3 -6y^2 -4y -15y -6y^2 +3y^3
sandra
  • sandra
and combine like terms (all of the y's, y^2 and y^3): -y^3 -12y^2 -19y
sandra
  • sandra
oops one second!
sandra
  • sandra
first one is 5y^3 heh, since 2+3=5, not -1 =)
sandra
  • sandra
so I get 5y^3 -12y^2 -19y

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