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Using factor theorem, factorize each of the following polynomial:

2y3 + y2 - 2y - 1

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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)

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