Answer:
[tex]2y^3 - 11y^2 + 12y + 9 = (y-3)(y-3)(2y+1) = (y-3)^2(2y+1)[/tex]
Step-by-step explanation:
We are given the expression:
[tex]2y^3 - 11y^2 + 12y + 9 = 0[/tex]
We have to factor this expression.
By hit and trial method, it was found that 3 is a root of given expression.
Thus, it can be written:
[tex]2y^3 - 11y^2 + 12y + 9 = (y-3)(2y^2 - 5y - 3)[/tex]
We can further factorize [tex]2y^2 - 5y - 3[/tex] by splitting the middle term technique.
[tex]2y^2 - 5y - 3 = 0\\=2y^2 - 6y + y - 3 = 0\\2y(y-3)+1(y-3) = 0\\(y-3)(2y+1)=0[/tex]
Thus, the expression can be factorized as:
[tex]2y^3 - 11y^2 + 12y + 9 = (y-3)(y-3)(2y+1) = (y-3)^2(2y+1)[/tex]
