To find the zeros of the polynomial function \( f(x) = x^3 - 40x^2 + 400x \), we can start by factoring the expression.
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First, we can factor out the common term from the polynomial: \[ f(x) = x(x^2 - 40x + 400) \]
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Next, we need to factor the quadratic \( x^2 - 40x + 400 \). We can either use the quadratic formula or complete the square. However, we can look for perfect squares first: \[ x^2 - 40x + 400 = (x - 20)^2 \] Therefore, we can rewrite the factored polynomial as: \[ f(x) = x(x - 20)^2 \]
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Now, to find the zeros, set \( f(x) = 0 \): \[ x(x - 20)^2 = 0 \]
This gives us: \[ x = 0 \quad \text{and} \quad (x - 20)^2 = 0 \implies x = 20 \]
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The zeros of the polynomial are:
- \( x = 0 \) (multiplicity 1)
- \( x = 20 \) (multiplicity 2)
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Now, we determine if the graph crosses or touches the x-axis at each zero:
- At \( x = 0 \) (multiplicity 1), since the multiplicity is odd, the graph crosses the x-axis.
- At \( x = 20 \) (multiplicity 2), since the multiplicity is even, the graph touches the x-axis and turns around.
Summary:
- The zeros are: \( 0, 20 \)
- Multiplicity:
- \( 0 \): multiplicity 1 (crosses the x-axis)
- \( 20 \): multiplicity 2 (touches and turns around at the x-axis)
Thus, the zeros are: \[ \text{The zero(s) is/are: } 0, 20 \]