Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(y)=2 y^{3}+y^{2}-2 y-1$

The constant term is 2

The factors of 2 are ± 1, ± 1/2

Let, y – 1= 0

=> y = 1

$f(1)=2(1)^{3}+(1)^{2}-2(1)-1$

= 2 + 1 – 2 – 1

= 0

So, (y – 1) is the factor of f(y)

Divide f(y) with (y – 1) to get other factors

By, long division

$2 y^{2}+3 y+1$

$y-1,2 y^{3}+y^{2}-2 y-1$

$2 y^{3}-2 y^{2}$

(-)     (+)

$3 y^{2}-2 y$

 

$3 y^{2}-3 y$

(-)       (+)

y – 1

y – 1

(-)   (+)

0

$=2 y^{3}+y^{2}-2 y-1=(y-1)\left(2 y^{2}+3 y+1\right)$

Now,

$2 y^{2}+3 y+1=2 y^{2}+2 y+y+1$

= 2y(y + 1) + 1(y + 1)

= (2y + 1) (y + 1) are the factors

Hence, $2 y^{3}+y^{2}-2 y-1=(y-1)(2 y+1)(y+1)$