<h2>Answer:</h2>
Option B.
2, multiplicity 1, crosses x-axis; 6, multiplicity 3, crosses x-axis
<h2>Explanations:</h2>
Given the function f(x) = 2(x - 2)(x - 6)^3
We need to get the zeros of the polynomial, its multiplicity and determine whether the graph crosses pr touches the x-axis
Get the zeros of the polynomial
Equating the function to zero will give;
2(x - 2)(x - 6)^3 = 0
This can be splitted as shown:
2(x - 2) = 0
x - 2 = 0
x = 2
Similarly,
(x - 6)^3 = 0
x - 6 = 0
x = 6
This shows that the zeros of the polynomial are 2 and 6
Get the multiplicity and <em>determine whether the graph crosses or touches the x-axis at each x-intercept</em>
First, you must note that zeroes with odd multiplicity cross through the x-axis, while zeroes with even multiplicity just touch the x-axis
For the factor (x-2), since the power of the linear factor is 1, hence its multiplicity is 1 showing that the graph crosses the x-axis at this point.
Similarly, for the other factor (x-6)^3, the multiplicity is 3 (Odd) showing that the graph crosses the x-axis at this point.
Based on the explanation above, the correct option will be Option B.
2, multiplicity 1, crosses x-axis; 6, multiplicity 3, crosses x-axis
so 
and 
. We can simplify that radicand down into a perfect square times a prime number, and that is 
. 4 is a perfect square that can be pulled out as a 2, so the length of x is 
, the choice on the top right.