A stack of squares has 170 squares in the bottom row, 156 in the second row from the bottom, 142 in the third row from the bottom and 128 in the fourth row from the bottom. How many squares will there be in the 11th row from the bottom ? - a table always makes these sums easier!)

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

one way is to plot 2 points and connect them which is what the answer choices seem to be doing

it seems to be using the y intercept and the slope

y=mx+b, m=slope, b=y-intercept (where the line hits the y axis)

slope=rise/run

y-intercept is where the line croses the y axis which is the y value where x=0

you seem to have given the wrong slope (you didn't put a number in front of te x, it should be 2/3 or 3/2 looking at the answer options)

y=? x-1

m=? because of your fault

b=-1

so it has a slope of ? and y intercept of -1, meaning it pases through (0,-1)

so draw point (0,-1) (so answer is one of the first 2 option)

if slope is 2/3, then go up 2 and right 3 then draw line through those 2 points

if slope is 3/2, then go up 3 and right 2 then draw line through those 2 points