5.(iv) Factorise :

   (iv)    2y^3 + y^2 - 2y - 1

Answers (1)

Given polynomial is  2y^3 + y^2 - 2y - 1

Now, by hit and trial method we observed that  (y-1)  is one of the factors of the given polynomial.

By long division method, we will get


We know that  Dividend = (Divisor \times Quotient) + Remainder

2y^3 + y^2 - 2y - 1= (y-1)(2y^2+3y+1)

                                       = (y-1)(2y^2+2y+y+1)
             
                                       = (y-1)(2y+1)(y+1)

Therefore, on factorization of  2y^3 + y^2 - 2y - 1 we will get  (y-1)(2y+1)(y+1)

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