To factor \( g(x) = x^3 - 3x^2 - 10x + 24 \) using the Factor Theorem, we start by recognizing that since \( -3 \) is a zero of \( g(x) \), \( g(-3) = 0 \). According to the theorem, this means that \( (x + 3) \) is a factor of \( g(x) \).
We will perform polynomial long division to divide \( g(x) \) by \( (x + 3) \):
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Set up the long division: \[ \begin{array}{r|rrrr} x + 3 & 1 & -3 & -10 & 24 \ & & 1 & 0 \ \hline
& 1 & 0 & -10 & 24 \ \end{array} \] -
Divide the leading term:
- Divide the leading term \( x^3 \) by \( x \), which gives \( x^2 \).
- Multiply \( x^2 \) by \( x + 3 \) to get \( x^3 + 3x^2 \).
- Subtract \( x^3 + 3x^2 \) from \( g(x) \): \[ g(x) - (x^3 + 3x^2) = (x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x^2 - 10x + 24. \]
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Repeat the process:
- Divide \( -6x^2 \) by \( x \) to get \( -6x \).
- Multiply \( -6x \) by \( x + 3 \) to get \( -6x^2 - 18x \).
- Subtract: \[ -6x^2 - 10x + 24 - (-6x^2 - 18x) = 8x + 24. \]
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Continue:
- Divide \( 8x \) by \( x \) to get \( 8 \).
- Multiply \( 8 \) by \( x + 3 \) to get \( 8x + 24 \).
- Subtract: \[ 8x + 24 - (8x + 24) = 0. \]
The result of the division is: \[ g(x) = (x + 3)(x^2 - 6x + 8). \]
Next, we will factor the quadratic \( x^2 - 6x + 8 \):
- We need to find two numbers that multiply to \( 8 \) and add to \( -6 \). The numbers \( -2 \) and \( -4 \) satisfy this: \[ x^2 - 6x + 8 = (x - 2)(x - 4). \]
Putting it all together, we have: \[ g(x) = (x + 3)(x - 2)(x - 4). \]
Thus, the complete factorization of \( g(x) \) into linear factors is: \[ \boxed{(x + 3)(x - 2)(x - 4)}. \]