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Answer:
Explanation:
Consider a zero at x=1.
If it has multiplicity 1, the corresponding factor is (x -1). Now consider the sign of that factor for x = 1.1. The value is (1.1 -1) = 0.1, a positive number.
When the value of x is 0.9, the factor is (0.9 -1) = -0.1, a negative number.
This tells you that values to the left of a zero make the factor(s) associated with that zero be negative. For x-values to the right of the zero, the factor(s) will be positive.
If the multiplicity is even, the product of those factors is positive on either side of the zero. (A negative number to an even power is a positive number.) That is, the graph will not change sign from one side of an even-multiplicity factor to the other. At an odd-multiplicity factor, the graph changes sign.
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For the problem at hand, the only even-multiplicity factor is (x -1), so the graph will not change sign at x=1. It will touch the axis (be tangent to the x-axis), but will not cross. The other zeros have odd multiplicity (1 or 3), so the graph will change sign there.
It is worthwhile to note that the higher the multiplicity, the "flatter" the graph is at the zero. You can see this in the attachment by comparing the shape of the graph at x=6 with the shape at x=2 (also an upward zero crossing).
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Additional comment
You may wonder why the curve is below the x-axis at x=1 when the product of the factors in that neighborhood is always positive. The reason is that there are an odd number of zeros to the right of x=1, so the product of those factors is negative.