Divide: (6y^2 - 35y + 36)/ (3y - 4) =?

Answer 1

(6y^2 - 35y + 36)=6y^2-27y-8y+36=3y(2y-9)-4(2y-9)=(3y-4)(2y-9).
(6y^2 - 35y + 36)/ (3y - 4)=2y-9 answer

Answer 2

i really have no idea how the two people above me got a 2y, but i can understand how they got -9. here is how i solved it, i divided it just like u said to:

(6y²-35y+36)/(3y-4) (with (3y-4) being the denominator):
so the 6y² & the 3y cancell each other out & become: 3y. the -35y doesnt have ne thing to cancell out with so it just stays the same. next the 36 & -4 cancell each out &
become: -9

so now ur left with 3y-35y-9.
(when ur dealing with variables in a math problem whether ur adding, subtracting, multiplying, or dividing, the biggest variable always goes first. i.e. 6z³+8z-3z. since the z on the 6 is cubed, the 6 goes first. it doesnt matter that the 8 is bigger than the 6. & yes, 35 is bigger than 3, but its a -35, therefore, it goes second)

so the answer is -32y-9.

Answer 3

= (3y - 4).(2y - 9) / (3y - 4)
= 2y - 9

Answer 4

(2y-9)