InsanelyChaotic
2x3 - 5x2y + xy2 + 2y3
a. Simplify (2x2 + y2)(x - 2y) = ?
b. Simplify (2x + y)(x2 -3xy + 2y 2 ) = ?
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timo86m
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where is c and d?
timo86m
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\[2\,{x}^{3}-5\,{x}^{2}y+x{y}^{2}+2\,{y}^{3}\]
did you mean that?
InsanelyChaotic
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yes
timo86m
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ooh where are c and d then :) i got my own answer can i see c and d?
InsanelyChaotic
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theyre not right for this
timo86m
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so far i think it is b but i feel it can be simplified further :)
InsanelyChaotic
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elimination
InsanelyChaotic
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its a
InsanelyChaotic
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;)
timo86m
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was it a?
timo86m
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well i get :)
(-y+x)*(x-2*y)*(y+2*x) fully factored
InsanelyChaotic
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idk
timo86m
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:) i'll check a and b :)
timo86m
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i dont think it is a :(
timo86m
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now for b
InsanelyChaotic
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its b!!!!!!!!!!!!!!!!!!:)))))))))))))))))))))))))))
timo86m
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told ya ;)
timo86m
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and i check to and it is b :D
timo86m
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If you know algrabeic long division then there is a theory of if you divide your given polynomial in this case
2x3 - 5x2y + xy2 + 2y3
by a suspected binomial or trinomial
(2x + y) or (x2 -3xy + 2y 2 )
^ those are from B the 2 expressions inside the ( )
and you get 0 then that trinomial or binomial is a factor and therfore part of the answer :)
timo86m
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if you get a remainder of 0 :P
timo86m
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in other words
2x3 - 5x2y + xy2 + 2y3
-------------------- = (x2 -3xy + 2y 2 ) with no remainder
(2x + y)
and obviously
2x3 - 5x2y + xy2 + 2y3
--------------------- = 2x+y with no remainder
(x2 -3xy + 2y 2 )
timo86m
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But if you dont know algabreic long division dont bother with that method :P