y^2 - 18y + 72?

Question

show work please

Answer 1

I'm guessing this is equal to zero, right?
y^2 - 18y +72 = 0

First, multiply the first and the last term:
1 * 72 = 72

Now you need two numbers that when multiplied give 72 and that when added together give you the second term which is -18. So I would say, 12 and 6.
-12*-6 = 72
-12 + -6 = 18

Now, split the second term into two:
(y^2 -12y)+(-6y +72) = 0

Now factorize:
y(y - 12) -6(y - 12) = 0

The terms in both parenthesis should always be equal to each other. If this is the case, you're on the right path. Now group the terms:
(y-6)(y-12) = 0

The first parenthesis has the terms that were outside the parenthesis and the second set of brackets has the terms that were already inside the parenthesis. Now just solve for y in each set of brackets:
y - 6 = 0
y = 6

And,
y - 12 = 0
y = 12

Plug this numbers into the original equation to see if you got the right answer. Hope this helps :)

Answer 2

y^2-18y+72=
(y - )(y - ) Because 72 is pos so both second terms must nbe neg

factors of 72 are: {(72, 1); (36, 2); (18, 4); (12, 6); (9,8) Only -12 -6 would give you -18y so:
(y-12)(y-6)

Answer 3

(y-6)(y-12)

y x y=y ^2
-6 x y=-6y
-12 x y=-12y
-6 x -12= 72

y^2-18y+72