1. The maximum area is when the rectangle is a square. So the three fenced sides are equal and have length 150/3=50 yd.

Let's look at the logic: the sides of the rectangle are L and W so the area, A, is LW and the perimeter, P=2L+2W. L=(P-2W)/2 and A=W(P-2W)/2=(1/2)(WP-2W^2). The maximum value of A is obtained by differentiating the quadratic with respect to W: (1/2)(P-4W)=0 at a turning point, so W=P/4 and L=(P-P/2)/2=P/4. Therefore L=W=P/4. The enclosed area is therefore a square, and we know that the three sides plus another unfenced length, L, make up the perimeter P=4L=150+L, so 3L=150 and L=50 yd.

Alternative solution avoiding calculus:

a. A=LW where L=length of rectangular area, W=width.

b. 150=2L+W.

c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L).

d. A=L(P-2L)/2=(LP-2L^2)/2=LP/2-L^2.

e. f(L)=-L^2+LP/2-A in standard quadratic form f(x)=ax^2+bx+c, where a=-1, b=P/2, c=-A and x=L.

f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area.

g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16.

From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd.

h. Maximum area is 50^2=2500 sq yd.

2. Synthetic division (1)

-3 | 2 -3 -45 -54

......2 -6..27...54

......2 -9 -18 | 0

3. Synthetic division (2)

6 | 2..-9 -18

.....2 12..18

.....2...3..| 0

This quotient is 2x+3, representing the third factor of the polynomial.

4. Synthetic division (3)

-4 | 2 -3 -45..-54

......2 -8..44.....4

......2 -11 -1 | -50

The Remainder Theorem tells us that substituting x=-4 into the polynomial gives us the same remainder. The substitution gives: -128-48+180-54=-50.

The zeroes are -3, 6 and -3/2, because synthetic division (2) tells us that 2x+3 is a factor.