To find the zeros of the function $f(x) = x^3 - 5x^2 + 6x$, we need to solve the equation $f(x) = 0$.

Step 1: Set up the equation

$x^3 - 5x^2 + 6x = 0$

Step 2: Factor out $x$

Since each term contains $x$, we can factor out $x$ from the expression: $x(x^2 - 5x + 6) = 0$

This gives us two factors:

1. $x = 0$
2. $x^2 - 5x + 6 = 0$

Step 3: Solve $x^2 - 5x + 6 = 0$

To solve $x^2 - 5x + 6 = 0$, we factor it further: $x^2 - 5x + 6 = (x - 2)(x - 3) = 0$

Step 4: Find the solutions

Setting each factor equal to zero gives us:

1. $x = 0$
2. $x - 2 = 0 \Rightarrow x = 2$
3. $x - 3 = 0 \Rightarrow x = 3$

Solution

The zeros of $f(x) = x^3 - 5x^2 + 6x$ are: $x = 0, \quad x = 2, \quad x = 3$

Would you like a detailed explanation of factoring or have any other questions?

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Related Questions:

1. How do you factor trinomials in quadratic form?
2. What are the zeros of the polynomial $g(x) = x^3 - 4x^2 + 4x$?
3. How can you verify the zeros of a polynomial?
4. What is the significance of multiplicity in the zeros of a polynomial?
5. How do the zeros of a polynomial relate to its graph?

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Tip: When finding zeros, always start by factoring out any common terms.