help me with math !!!!?

Question

-2y(y+2) = 0


( x - 1)( x +2) = 0

x^ - 4 = 0

Answer 1

- 2y(y + 2) = 0

- (2y² + 4y = 0

- 2y² - 4y = 0

- - - - - - - - -

FOIL

(X - 1)(X + 2) =

x² - x + 2x - 2 =

x² + x - 2

- - - - - - - - -s-

Answer 2

Answer 1:
-2y(y + 2) = 0

Either -2y = 0
or y + 2 = 0

If -2y = 0, y = 0
If y + 2 = 0, y = -2

y = 0, -2 are the two possible values of y

Answer 2:
(x - 1)(x + 2) = 0

Either x - 1 = 0
or x + 2 = 0

x = 1, -2 are the solutions of this quadratic

Answer 3

-2y(y+2) = 0
– 2 y^2 = 0 or ( y + 2) = 0
Therefore y = 0 or y = – 2

Similarly
( x - 1)( x +2) = 0
( x - 1) = 0 or ( x +2) = 0
Therefore x = 1 or x = – 2

Answer 4

You can divide the equation by -2 to get:

y(y+2) = 0

So y=0 or y = -2

b)Either x - 1 = 0 (x = 1)
or x + 2 = 0 (x = -2)

c)

Answer 5

Q1) Change it to quadratics first.

-2y(y+2)=0 = -2y^2 - 4y=0

After putting the numbers into the quadratic formula, you will finally arrive at the answer y=0 or y=-2

Q2) x=1 or x=-2

Answer 6

divide both sides by 2
s
y(y+2)=0
thus either y=0 or y+2=0
thus y=0 or y=-2

similarly
x=1 or x=-2

Answer 7

I would assume you need to solve those equations for x and y, right?
You need to know that the product is zero when either of the factors is zero.

1. y is zero or y+2 is zero
solution: y =0, y= -2

2. x-1 is zero or x+2 is zero
solution: x = 1 or x = -2

3. x^2 - 4 = (x-2)(x+2)=0
x = 2, or x = -2

Answer 8

-2y(y+2)=0
or, -y2 - 2y = 0
or,y(-y-2)=0
so either y=0 or y=-2

(x-1)(x+2)=0
then either x = 1 o x =-2

Answer 9

exactly what is it you're trying to do?