# Using factor theo

Let, f (y) = 2y3+y2-2y-1

The factors of the constant term - 1 are

The factor of the coefficient of y3 is 2. Hence, possible rational roots are

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)

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