Answer: answer is in picture. :)
Step-by-step explanation:
The answer is C I just did this
Answer:
B. 14x^3+39x^2+18x+20
Step-by-step explanation:
Given polynomials are:

The product of given polynomials is:
14x^3+39x^2+18x+20
Hence, Option B is correct ..
<h3>
Answers:</h3>

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Work Shown:
Part 1

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.
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Part 2

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2
In this case, a = sqrt(x-1) and b = 5.
You could also use the box method as a visual way to expand out 
Answer:
B
Step-by-step explanation:
Points A,B ,C are all to the left of 0 so they are negative
We want a point -3 1/2
The points A and B are to the left of -3 so they are more negative than -3
A is at -5 so it cannot be -3 1/2