2y^3-10y^2+12y...need to factor again..i got to the second step...
2y(y^2-5y+6)...think that's right but i don't get how to factor that out..help plz..
Go for one more factoring step and you get
2 y (y-2)(y-3)
Try it this way...
2y^3 - 10y^2 + 12y
Switch A&C to make it easier (always acceptable, just remember to keep your signs):
12y - 10y^2 + 2y^3
Factor out a GCF, in this case it's 2y:
2y(6 - 5y + y^2)
Next, factor a trinomial:
2y(2 - 1y)(3 - 1y)
This trinomial is your final answer.
Hope this helps you!
To factor the expression 2y^3 - 10y^2 + 12y further, you correctly pulled out the greatest common factor, which is 2y. This gives you the expression 2y(y^2 - 5y + 6).
To factor the quadratic expression y^2 - 5y + 6, you need to find two numbers that multiply to 6 and add up to -5. The numbers that satisfy this are -2 and -3, because (-2) * (-3) = 6 and (-2) + (-3) = -5.
Now, we can rewrite the quadratic expression as (y - 2)(y - 3).
Therefore, factoring the original expression completely, we have:
2y(y^2 - 5y + 6) = 2y(y - 2)(y - 3).
You have factored the expression correctly!