Factorise 2y3+y2−2y−1

1.	Factorise 2y3+y2−2y−1
Answer» $\huge\boxed{\underline{\mathcal{\red{A} \green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$ 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) ${\bf{\orange{FOLLOW ME}}}$

Answer»

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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)