Question 4 (essay worth 10 points) (08.01, 08.02, 08.03 hc) create a factorable polynomial with a gcf of 3x. rewrite that polynomial in two other equivalent forms. explain how each form was created.

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.

asnwer is (g o h o f)(x) = 2x - 6/x + 7

step-by-step explanation:

given,

f(x) = x - 3/x

g(x) = x + 3

h(x) = 2x + 1

we need to find : (g o h o f)(x)

we can write (g o h o f)(x) = g[ h{ f(x) } ]

so first we find h{f(x)}

h{f(x)} = f(x) + 3 = (x - 3/x) + 3

h{f(x)} = x - 3/x + 3

now we find g[ h{ f(x) } ].

g[ h{ f(x) } ] = g(x - 3/x + 3)       {plug h{f(x)} = x - 3/x + 3 in g[h{f(x)}] }

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

                    = 2x - 6/x + 6 + 1

                    = 2x - 6/x + 7

hence we get (g o h o f)(x) = 2x - 6/x + 7