Question

factorise 2y3+y2−2y−1​

Answer 1

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Step-by-step explanation:

Let p(y)=2y³+y²-2y-1

, 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)

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Answer 2

Answer:

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

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

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

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

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

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

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

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