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write the following quotient in simplest form 3y^3 + 2y^2 - 7y + 2 / 3y-1
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You can factorise \[3y^3+2y^2-7y+2\]
putting y=1, this expressions goes to 0, so (y-1) is a factor dividing the numerator by (y-1) you get \[3y^2+5y-2\]and this gets factorised into (3y-1)(y+2)
So the whole expression becomes:\[\frac{ (y-1)(3y-1)(y+2) }{3y-1 }\]from which we get:\[(y-1)(y+2)\]
so once u divide and you then get (y-1)(y+2)
Yes you are correct
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