Question

vi. (y² + 2y + 2) (y² - 2y + 2) ​

Answer 1

Answer:

y

2

+3y−y−3=0

⇒y(y+3)−1(y+3)=0

taking (y+3)common,

⇒(y+3)(y−1)=0

∴ y+3=0 OR y−1=0

y=−3 OR y=1