To find the zeros of the polynomial function \( n(x) = -0.5x^3 - 1.5x^2 + 2x \), we need to set the function equal to zero and solve for \( x \):
\[ -0.5x^3 - 1.5x^2 + 2x = 0 \]
We can factor out \( -0.5x \):
\[ -0.5x(x^2 + 3x - 4) = 0 \]
Setting each factor equal to zero gives us:
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\( -0.5x = 0 \) \[ x = 0 \]
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For the quadratic \( x^2 + 3x - 4 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = 3 \), and \( c = -4 \).
Calculating the discriminant: \[ b^2 - 4ac = 3^2 - 4(1)(-4) = 9 + 16 = 25 \]
Now substituting into the formula: \[ x = \frac{-3 \pm \sqrt{25}}{2 \cdot 1} = \frac{-3 \pm 5}{2} \]
This gives us two more roots: \[ x = \frac{2}{2} = 1 \quad (\text{using the } + \text{ sign}), \] \[ x = \frac{-8}{2} = -4 \quad (\text{using the } - \text{ sign}). \]
Thus, the zeros of \( n(x) \) are \( x = -4 \), \( x = 0 \), and \( x = 1 \).
In order from least to greatest, the zeros are:
\[ \text{The zeros of } n(x) \text{ are } x = -4, x = 0, \text{ and } x = 1. \]
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