Use the Factor Theorem to factor g(x)=x^3−3x^2−10x+24 into a product of linear factors, given that −3 is a zero of ​g(x)

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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 \( (x + 3) \).

  1. Polynomial Long Division:

    Divide \( g(x) \) by \( (x + 3) \):

    \[ x^3 - 3x^2 - 10x + 24 \div (x + 3) \]

    • Divide the leading term: \( \frac{x^3}{x} = x^2 \).

    • Multiply \( (x + 3) \) by \( x^2 \): \[ x^2(x + 3) = x^3 + 3x^2 \]

    • Subtract this from \( g(x) \): \[ (x^3 - 3x^2 - 10x + 24) - (x^3 + 3x^2) = -6x^2 - 10x + 24 \]

    • Now, divide the leading term \( -6x^2 \) by \( x \): \[ -6x^2 \div x = -6x \]

    • Multiply \( (x + 3) \) by \( -6x \): \[ -6x(x + 3) = -6x^2 - 18x \]

    • Subtract: \[ (-6x^2 - 10x + 24) - (-6x^2 - 18x) = 8x + 24 \]

    • Finally, divide \( 8x \) by \( x \): \[ 8x \div x = 8 \]

    • Multiply \( (x + 3) \) by \( 8 \): \[ 8(x + 3) = 8x + 24 \]

    • Subtract: \[ (8x + 24) - (8x + 24) = 0 \]

The result of the division is \( x^2 - 6x + 8 \), and there is no remainder. Therefore, we have:

\[ g(x) = (x + 3)(x^2 - 6x + 8) \]

  1. 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:

\[ x^2 - 6x + 8 = (x - 2)(x - 4) \]

  1. 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:

\[ g(x) = (x + 3)(x - 2)(x - 4) \]