Answer:
The probably genotype of individual #4 if 'Aa' and individual #6 is 'aa'.
Step-by-step explanation:
In a non sex-linked, dominant trait where both parents carry and show the trait and produce children that both have and don't have the trait, they would each have a genotype of 'Aa' which would produce a likelihood of 75% of children that carry the dominant traint and 25% that don't. Since the child of #1 and #2, #5, does not exhibit the trait, nor does the significant other (#6), then they both must have the 'aa' genotype. However, since #4 displays the dominant trait received from the parents, it is more likely they would have the 'Aa' genotype as by the punnet square of 'Aa' x 'Aa', 50% of their children would have the 'Aa' phenotype.
For this case we have the following arithmetic sequence:

To write in function form, we apply distributive property to the terms within parentheses:

Different signs are subtracted and the major sign is placed.
We simplify:

Answer:

Option C
This is the concept of polynomials, to get the relationship between the (x-3) and (x^3+4x^2+2) we shall proceed as follows;
we first test if (x-3) is a polynomial;
x-3=0
thus;
x=3
substituting the value of x in the polynomial we get:
(3)^3+4(3)^2+2
=65
since the result is not equal to 0, we conclude that there is no relationship between (x-3) and the polynomial. The answer is B. (X-3) is not a factor
Add them all up and divide by 6
160/6 = 26.7