To factor the polynomial \( g(x) = x^3 - 3x^2 - 10x + 24 \) using the Factor Theorem, we know that since \( -3 \) is a zero of \( g(x) \), \( (x + 3) \) is a factor of \( g(x) \).
We can use polynomial long division to divide \( g(x) \) by \( (x + 3) \).
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Polynomial Long Division:
Divide \( g(x) \) by \( (x + 3) \):
\[ x^3 - 3x^2 - 10x + 24 \div (x + 3) \]
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Divide the leading term: \( \frac{x^3}{x} = x^2 \).
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Multiply \( (x + 3) \) by \( x^2 \): \[ x^2(x + 3) = x^3 + 3x^2 \]
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Subtract this from \( g(x) \): \[ (x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x^2 - 10x + 24 \]
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Now, divide the leading term \( -6x^2 \) by \( x \): \[ -6x^2 \div x = -6x \]
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Multiply \( (x + 3) \) by \( -6x \): \[ -6x(x + 3) = -6x^2 - 18x \]
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Subtract: \[ (-6x^2 - 10x + 24) - (-6x^2 - 18x) = 8x + 24 \]
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Finally, divide \( 8x \) by \( x \): \[ 8x \div x = 8 \]
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Multiply \( (x + 3) \) by \( 8 \): \[ 8(x + 3) = 8x + 24 \]
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Subtract: \[ (8x + 24) - (8x + 24) = 0 \]
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The result of the division is \( x^2 - 6x + 8 \), and there is no remainder. Therefore, we have:
\[ g(x) = (x + 3)(x^2 - 6x + 8) \]
- Factoring \( x^2 - 6x + 8 \):
Next, we can factor \( x^2 - 6x + 8 \). We look for two numbers that multiply to \( 8 \) and add to \( -6 \). These numbers are \( -2 \) and \( -4 \).
Thus, we can write:
\[ x^2 - 6x + 8 = (x - 2)(x - 4) \]
- Final Factored Form:
Now, combining everything, we have:
\[ g(x) = (x + 3)(x - 2)(x - 4) \]
Therefore, the complete factorization of \( g(x) \) into linear factors is:
\[ g(x) = (x + 3)(x - 2)(x - 4) \]