Given that x = 2 is a zero of f(x)=x3−3x2−10x+24

To factor the polynomial \( f(x) = x^3 - 3x^2 - 10x + 24 \) given that \( x = 2 \) is a zero, we can use the Factor Theorem, which states that if \( c \) is a zero of the polynomial \( f(x) \), then \( (x - c) \) is a factor of \( f(x) \).

Since \( x = 2 \) is a zero, we know that \( (x - 2) \) is a factor of \( f(x) \).

Next, we can perform polynomial long division to divide \( f(x) \) by \( (x - 2) \):

1. Divide the leading term: \( \frac{x^3}{x} = x^2 \).
2. Multiply \( (x - 2) \) by \( x^2 \): \( x^2(x - 2) = x^3 - 2x^2 \).
3. Subtract from \( f(x) \): \[ (x^3 - 3x^2 - 10x + 24) - (x^3 - 2x^2) = -x^2 - 10x + 24. \]
4. Next, divide the leading term again: \( \frac{-x^2}{x} = -1 \).
5. Multiply \( (x - 2) \) by \( -1 \): \( -1(x - 2) = -x + 2 \).
6. Subtract: \[ (-x^2 - 10x + 24) - (-x + 2) = -9x + 22. \]
7. Divide again: \( \frac{-9x}{x} = -9 \).
8. Multiply \( (x - 2) \) by \( -9 \): \( -9(x - 2) = -9x + 18 \).
9. Subtract: \[ (-9x + 22) - (-9x + 18) = 4. \]

Since we only have a constant, this indicates that the result of the polynomial division is \( x^2 - x - 12 \), and we can write: \[ f(x) = (x - 2)(x^2 - x - 12). \]

Now, we need to factor \( x^2 - x - 12 \): To factor \( x^2 - x - 12 \), we look for two numbers that multiply to \(-12\) and add up to \(-1\). The numbers \(-4\) and \(3\) fit since: \[ -4 + 3 = -1 \quad \text{and} \quad -4 \cdot 3 = -12. \] Thus, we can write: \[ x^2 - x - 12 = (x - 4)(x + 3). \]

Putting it all together, we have: \[ f(x) = (x - 2)(x - 4)(x + 3). \]

So, the correct factorization is: \[ \boxed{f(x) = (x - 2)(x + 3)(x - 4)}. \]