1. If a polynomial of Degree 4 has an imaginary zero at -2i and a real zero at 5, how many imaginary zeros does it have? and how many real zeros does it have?

2. If a polynomial has a Degree of two, how many imaginary zeros could it have? and how many real zeros could it have?

Complex roots for polynomials equations will always come in conjugate pairs.  Since 1-2i is a root, 1+2i will also be a root.  (Just switch the sign of the imaginary part to its opposite.)

Step-by-step explanation:

So far we have 1-2i and 1+2i as roots.  We need two more to make a 4th degree polynomial.

x=1 with multiplicity of 2 means that we count the root twice when we write down the polynomial: (x - 1)(x - 1) = (x - 1)2

Putting it all together, we can construct our polynomial of degree 4 as

y = f(x) = (x - 1)(x - 1)[x - (1-2i)][x - (1+2i)]

Multiply the factors (FOIL):

[x - (1-2i)][x - (1+2i)] =

x2 - (1+2i)x - (1-2i)x + (1-2i)(1+2i) =

x2 - x - 2ix - x + 2ix + (1 + 2i - 2i + 4) =

x2 - 2x + 5

(x - 1)(x - 1) = x2 - x - x + 1 = x2 - 2x + 1

Now multiply the two underlined expressions:

y = (x2 - 2x + 5)(x2 - 2x + 1)

= x4 - 2x3 + 5x2 - 2x3 + 4x2 - 10x + x2 - 2x + 5  by multiplying term by term

= x4 - 4x3 + 10x2 -12x + 5 by collecting like terms.

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

answer: i believe the answer is e

step-by-step explanation: