Question

solve it ....... 3y² + 2y + 1​

Answer 1

Answer:

(3y + 1)(y-1)

Explanation:

3y^2 - 2y - 1

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

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

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

Answer 2

There is no value of y which will Satisfy the condition 3y^2+2y+1=0

so we can say

y belong to ø

Reason:-

Here Discriminant=b^2-4ac

b^2-4ac=4-4×3×1

=-8, which is non negative

and we know that if D<0 then the root of the equation will be imaginary.