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