To find the zeros of the polynomial function \( n(x) = -0.5x^3 - 1.5x^2 + 2x \), we set the function equal to zero and solve for \( x \):
\[ -0.5x^3 - 1.5x^2 + 2x = 0 \]
Factoring out a common term, we can simplify this:
\[ x(-0.5x^2 - 1.5x + 2) = 0 \]
From this, we have one zero:
\[ x = 0 \]
Next, we need to solve the quadratic equation:
\[ -0.5x^2 - 1.5x + 2 = 0 \]
To make computations easier, we can multiply through by -2 to eliminate the fraction:
\[ x^2 + 3x - 4 = 0 \]
Now we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 3 \), and \( c = -4 \):
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 + 16}}{2} = \frac{-3 \pm \sqrt{25}}{2} \]
\[ x = \frac{-3 \pm 5}{2} \]
Calculating the two potential values for \( x \):
- \( x = \frac{-3 + 5}{2} = \frac{2}{2} = 1 \)
- \( x = \frac{-3 - 5}{2} = \frac{-8}{2} = -4 \)
Thus, the zeros of the polynomial \( n(x) \) are:
- \( x = -4 \)
- \( x = 0 \)
- \( x = 1 \)
In order from least to greatest, the zeros are:
\[ x = -4, \quad x = 0, \quad x = 1 \]
So you can write:
The zeros of \( n(x) \) are \( x = -4 \), \( x = 0 \), and \( x = 1 \).
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