To express $g(x) = x^3 + x^2 - 10x - 12$ as a product of linear factors, given that $-3$ is a zero, we follow these steps:

Step 1: Perform synthetic or polynomial division by $x + 3$

Since $-3$ is a zero, $x + 3$ is a factor of $g(x)$. Using synthetic or polynomial division, divide $g(x)$ by $x + 3$.

Synthetic Division

The coefficients of $g(x)$ are $[1, 1, -10, -12]$. Perform division:

1 | 1 1 -10 -12
-3 | -3 6 12

 1   -2   -4     0  

The quotient is $x^2 - 2x - 4$ with a remainder of 0 (as expected since $-3$ is a zero).

Thus, we can write: $g(x) = (x + 3)(x^2 - 2x - 4)$

Step 2: Factorize $x^2 - 2x - 4$ further

We now factorize $x^2 - 2x - 4$ using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad \text{where } a = 1, b = -2, c = -4.$ $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2} = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}.$

Thus, $x^2 - 2x - 4 = (x - (1 + \sqrt{5}))(x - (1 - \sqrt{5}))$.

Final Answer

Expressing $g(x)$ as a product of linear factors: $g(x) = (x + 3)(x - (1 + \sqrt{5}))(x - (1 - \sqrt{5})).$

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Related Questions:

1. How is synthetic division used to factor polynomials efficiently?
2. What steps ensure accurate factoring when a zero is known?
3. How does the quadratic formula assist in finding real or complex roots?
4. What are linear factors, and why are they useful for polynomial analysis?
5. Can complex numbers be part of the linear factorization?

Tip:

Always verify your factored form by expanding it back to ensure it matches the original polynomial.