Given that \( x = 2 \) is a zero of the polynomial \( f(x) = x^3 - 3x^2 - 10x + 24 \), we can apply the Factor Theorem, which states that if \( c \) is a root of the polynomial \( f(x) \), then \( f(x) \) can be factored as \( f(x) = (x - c)g(x) \) for some polynomial \( g(x) \).
Since \( x = 2 \) is a zero, we can factor \( f(x) \) as:
\[ f(x) = (x - 2)g(x) \]
To find \( g(x) \), we can perform polynomial long division of \( f(x) \) by \( (x - 2) \):
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Divide \( x^3 \) by \( x \) to get \( x^2 \).
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Multiply \( (x - 2) \) by \( x^2 \) to get \( x^3 - 2x^2 \).
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Subtract \( (x^3 - 2x^2) \) from \( f(x) \):
\[ (x^3 - 3x^2 - 10x + 24) - (x^3 - 2x^2) = -3x^2 + 2x^2 - 10x + 24 = -x^2 - 10x + 24 \]
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Divide \( -x^2 \) by \( x \) to get \( -x \).
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Multiply \( (x - 2) \) by \( -x \) to get \( -x^2 + 2x \).
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Subtract:
\[ (-x^2 - 10x + 24) - (-x^2 + 2x) = -10x - 2x + 24 = -12x + 24 \]
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Divide \( -12x \) by \( x \) to get \( -12 \).
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Multiply \( (x - 2) \) by \( -12 \) to get \( -12x + 24 \).
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Subtract:
\[ (-12x + 24) - (-12x + 24) = 0 \]
Thus, we have:
\[ f(x) = (x - 2)(x^2 - x - 12) \]
Now we can factor the quadratic \( x^2 - x - 12 \). We are looking for two numbers that multiply to -12 and add to -1; these numbers are -4 and 3. So we can factor it as follows:
\[ x^2 - x - 12 = (x - 4)(x + 3) \]
Thus, we have:
\[ f(x) = (x - 2)(x - 4)(x + 3) \]
The final factorization of \( f(x) \) into linear factors is:
\[ f(x) = (x - 2)(x - 4)(x + 3) \]
The correct response from your list is:
f(x) = ( x - 2)(x + 3)(x - 4)
1 answer