# 5.(iv) Factorise :   (iv)    $2y^3 + y^2 - 2y - 1$

G Gautam harsolia

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  $(y-1)(2y+1)(y+1)$

Exams
Articles
Questions