What is a counterexample for the following conjecture ? All four -sided figures are squares.

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.