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1)A.f(x)= x^2+4
B.f(x)= x^2-4x^2+4x-16
C.f(x)= x^2+4x^2+4x+16
D.f(x)= x^2-4x^2-4x+16


4)I don't know what they mean either but this is all it says.

what do they mean when they say like 2i,what is that?

1)Write a polynomial function of least degree with integral coefficients whose zeros include 4 and 2i.
answer= f(x)= x^2-4x^2+4x-16

2)List all of the possible rational zeros of f(x)= 3x^3-2x^2+7x+6
dont know

3)Find all of the rational zeros of f(x)= 4x^3-3x^2-22x-15
dont know

4)Find (f.g)(x) for f(x)= 3x^2 and g(x)= 5-x
answer= 3x^2-15x^2

1. imaginary roots always come in pairs, like in ±2i
so the factors would be (x+2i)(x-2i)(x-4)
expand it and you will have your answer.

2. I tried the factor theorem hoping for some f(a)=0 where a=±1,±2,±3
None worked so I don't know how you are expected to do that one.
You could try ±1/3,±2/3 but that seems a bit too farfetched from the type of questions you seem to have

3. try f(-1) it will be a zero
so x+1 is a factor. Do synthetic division or long division, you should get an answer of 4x^2 - 7x - 15 which factors again.
(see if you can get zeros at x=-1,3,-5/4

4. I don't know if your textbook defines
(f∙g) as f(g(x)) or g(f(x)).

f(g(x)) = f(5-x) = 3(5-x)^2
= 75 - 30x + 3x^2

g(f(x)) = g(3x^2) = 5 - 3x^2

i is the symbol for √(-1), which is the imaginary unit number.
so 2i is really 2√(-1)

in other words i^2 = -1

eg. solve x^2 + 9=0
x^2 = -9
x = ±(√9)(√(-1))
x = ±3√(-1)
x = ±3i

1. After I expanded my answer to #1 above I got x^3 - 4x^2 + 4x - 16 which is the same as B if your first term is x^3, as it should be for all of those answers.

2. for this questions according to the answers I can see that they simply wanted those values that your would try in your f(x) function. As I noted in my first answer, the correct answer would appear to be C

4. So it is B
Look at my first solution how I got that.

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