Question

factorise 2y^3+y^2-2y-1

Answer 1

Answer:

2y³+y²-2y-1=(y-1)(y+1)(2y+1)

Step-by-step explanation:

Let p(y)=2y³+y²-2y-1

By trail , we find that,

p(1) = 2(1)³+1²-2(1)-1

= 2+1-2-1

= 0

By Factor theorem,

(y-1) is a factor of p(y).

Now,

2y³+y²-2y-1

=2y²(y-1)+3y(y-1)+1(y-1)

=(y-1)(2y²+3y+1)

=(y-1)(2y²+2y+1y+1)

=(y-1)[2y(y+1)+1(y+1)]

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

Therefore,

2y³+y²-2y-1=(y-1)(y+1)(2y+1)

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Answer 2

Given that,

An expression $2y^3+y^2-2y-1$

To find,

Factors of the above expression.

Solution,

We have,

$2y^3+y^2-2y-1$

Taking y² common from first two terms and -1 common from last two terms

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

Now using identity, $(a^2-b^2)=(a-b)(a+b)$

So,

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

So, the factors of $2y^3+y^2-2y-1$ is  $(y-1)(2y+1)(y+1)$.