Question

Hey

Divide 2y²+y²-2y-1 with (y-1)

That's all for now ^^"

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Explanation required..​

Answer 1

Answer



For doing this sum we have 2 procedure

First process is above

second one is below

y - 1 = 0

=> y = 0 + 1 = 1

p(y) = 2y² + y² - 2y -1

After substituting the value of y

p(-1) = 2(1)² + (1)² - 2(1) - 1

=> 2(1) + 1 - 2 - 1

=> 2 + 1 - 2 - 1

=> 0

Hope it helps you :)

Thanks :)

Answer 2

${\huge {\mathfrak {Answer :-}}}$

___________

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

= 3y^2 - 2y - 1

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

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

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

___________

Now,

[ 2y^2 + y^2 - 2y - 1 ] ÷ (y-1)

= (3y + 1)(y - 1) ÷ (y - 1)

= 3y + 1

___________

Thanks..