Question

factorise:
plz I will mark u brainliest

Answer 1

The question is wrong
Note it
4y^3-12y^2-y+3=0

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(((4 • (y3)) - (22•3y2)) - y) + 3

Step  2  :

Equation at the end of step  2  :

((22y3 - (22•3y2)) - y) + 3

Step  3  :

Checking for a perfect cube :

 3.1    4y3-12y2-y+3  is not a perfect cube 

Trying to factor by pulling out :

 3.2      Factoring:  4y3-12y2-y+3 

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  -y+3 
Group 2:  4y3-12y2 

Pull out from each group separately :

Group 1:   (-y+3) • (1) = (y-3) • (-1)
Group 2:   (y-3) • (4y2)
               -------------------
Add up the two groups :
               (y-3)  •  (4y2-1) 


Factorization is :       (2y + 1)  •  (2y - 1) 

Final result :

(2y + 1) • (2y - 1) • (y - 3)

y = - 1/2 ; 1/2 ; 3

This is bit complicated sorry for that

Hope this helps you

Please mark as brainliest