Question

factorise 2y^3+y^2-2y-1 ​

Answer 1

Factorise

2y3 + y2 - 2y - 1

2y3 + y2 - 2y - 1

Let p(y) = 2y3 + y2 - 2y - 1

By trial, we find that

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

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

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

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Here we have :-

p(y) = 2y³ + y² - 2y - 1

Constant = -1

$\Large{\boxed{\sf\:{Numerical\;value\;of\;(-1) = 1}}}$

All factors of 1 are 1 and -1

Hence,

p(-1) = 2(-1)³ + (-1)² - 2(-1) - 1

= -2 + 1 + 2 - 1

= 0

Therefore,

$\Large{\boxed{\sf\:{(y + 1)\;is\;factor\;of\;p(y)}}}$

Now,

2y³ + y² - 2y - 1 = 2y³ + 2y² - y² - y - y - 1

= (2y³ + 2y²) - (y² + y) - (y + 1)

= 2y²(y + 1) - y(y + 1) - 1(y + 1)

= (y + 1)(2y² - y - 1)

= (y + 1)(2y² - 2y + y - 1)

= (y + 1)[(2y² - 2y) + 1(y - 1)]

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

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