Let, f (y) = 2y3 + y2 - 2y - 1
The factors of the constant term - 1 are \(\pm\) 1
The factor of the coefficient of y3 is 2.
Hence,
possible rational roots are \(\pm1\pm\frac{1}{2}\)
We have,
f (1) = 2 (1)3 + (1)2 – 2 (1) - 1
= 2 + 1 – 2 - 1
= 0
So,
(y - 1) is a factor of f (y)
Let us now divide
f (y) = 2y3 + y2 - 2y - 1 by (y - 1) to get the other factors of f (x)
Using long division method, we get
2y3 + y2 - 2y - 1 = (y - 1) (2y2 + 3y + 1)
2y2 + 3y + 1 = 2y2 + 2y + y + 1
= 2y (y + 1) + 1 (y + 1)
= (2y + 1) (y + 1)
Hence,
2y3 + y2 - 2y - 1 = (y - 1) (2y + 1) (y + 1)