To factor the polynomial \( g(x) = x^3 - 3x^2 - 10x + 24 \) using the Factor Theorem, we will first confirm that \( -3 \) is a zero of \( g(x) \).
Substituting \( x = -3 \):
\[ g(-3) = (-3)^3 - 3(-3)^2 - 10(-3) + 24 \] \[ = -27 - 3(9) + 30 + 24 \] \[ = -27 - 27 + 30 + 24 \] \[ = -54 + 54 = 0 \]
Since \( g(-3) = 0 \), \( (x + 3) \) is a factor of \( g(x) \).
Next, we perform polynomial long division to divide \( g(x) \) by \( (x + 3) \):
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Divide the leading term of \( g(x) \) by the leading term of \( (x + 3) \): \[ \frac{x^3}{x} = x^2 \]
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Multiply \( x^2 \) by \( (x + 3) \): \[ x^2 (x + 3) = x^3 + 3x^2 \]
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Subtract: \[ g(x) - (x^3 + 3x^2) = (x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x^2 - 10x + 24 \]
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Next, divide the leading term of the new polynomial by the leading term of \( (x + 3) \): \[ \frac{-6x^2}{x} = -6x \]
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Multiply \( -6x \) by \( (x + 3) \): \[ -6x(x + 3) = -6x^2 - 18x \]
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Subtract again: \[ -6x^2 - 10x + 24 - (-6x^2 - 18x) = (-10x + 18x) + 24 = 8x + 24 \]
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Now, divide the leading term of \( 8x + 24 \) by the leading term of \( (x + 3) \): \[ \frac{8x}{x} = 8 \]
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Multiply \( 8 \) by \( (x + 3) \): \[ 8(x + 3) = 8x + 24 \]
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Subtract: \[ 8x + 24 - (8x + 24) = 0 \]
Now we see that:
\[ g(x) = (x + 3)(x^2 - 6x + 8) \]
Next, we will factor \( x^2 - 6x + 8 \) further. We need to find two numbers that multiply to \( 8 \) and add to \( -6 \). Those numbers are \( -2 \) and \( -4 \). Thus, we can factor \( x^2 - 6x + 8 \) as:
\[ x^2 - 6x + 8 = (x - 2)(x - 4) \]
Therefore, the complete factorization of \( g(x) \) is:
\[ g(x) = (x + 3)(x - 2)(x - 4) \]
So, the final answer is:
\[ g(x) = (x + 3)(x - 2)(x - 4) \]