2y³+y²-2y-1
Using the factor theorem:
If r is a root then (y-r) is a factor
The candidates for the roots are ± the factors each of whose
numerator is a factor of 1, the constant term, and whose
denominator is a factor of 2, the coefficient of the term
with the largest power of the variable, 2y³.
The candidates for roots are ±1, 
We try 1 using synthetic division:
1|2 1 -2 -1
| 2 3 1
2 3 1 0
The remainder is 0, so 1 is a root and we have now
factorised 2y³+y²-2y-1 as
(y-1)(2y²+3y+1)
We now factorise the trinomial 2x²+3x+1 as (2x+1)(x+1)
(y-1)(2y+1)(y+1)
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Without using the factor theorem:
2y³+y²-2y-1
Factorise out y² from the first two terms
Factorise out -1 from the last two terms
y²(2y+1)-1(2y+1)
Factorise out (2y+1)
(2y+1)(y²-1)
Factorise y²-1 as the difference of squares (y-1)(y+1)
(2y+1)(y-1)(y+1)
Edwin
