Which of the following circumstances would likely make completing the square th best method for solving a quadratic equation?
The leading coefficient is 1 and the linear term coefficient is an even number
The leading coefficient is not 1 and the constant is large
The difference of 2 perfect squares
The difference of 2 perfect cubes

The leading coefficient is 1 and the linear term coefficient is an even number

Step-by-step explanation:

Leading coefficient must be one for the application of completing square method. If the completing square is not one than we must apply other methods like Ac method, quadratic formula to solve the equation. To solve the equation with completing square the numerical value must be on other side of the equation. In this method we multiply linear term’s coefficient with one-half and square it therefore, completing square is suitable method for the equation having leading coefficient is 1 and linear coefficient is even.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

not sure but it may be b, as i said i'm not sure.