For the polynomial below, 1 is a zero. \[ h(x)=x^{3}-3 x^{2}-2 x+4 \] Express \( h(x) \) as a product of linear factors. \( h(x)=\square \)

We are given \[ h(x) = x^3 - 3x^2 - 2x + 4 \] and we know that \( x = 1 \) is a zero of \( h(x) \), which means \( (x-1) \) is a factor. **Step 1. Divide by \((x-1)\):** Using synthetic division with \( x = 1 \) and coefficients \( 1, -3, -2, 4 \): - Bring down the first coefficient: \( 1 \). - Multiply \( 1 \) by \( 1 \) and add to \(-3\): \( -3 + 1 = -2 \). - Multiply \( -2 \) by \( 1 \) and add to \(-2\): \( -2 + (-2) = -4 \). - Multiply \( -4 \) by \( 1 \) and add to \( 4 \): \( 4 + (-4) = 0 \). Thus, the quotient is the quadratic: \[ x^2 - 2x - 4. \] **Step 2. Factor the quadratic \( x^2 - 2x - 4 \):** To find its zeros, we use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \( a = 1 \), \( b = -2 \), and \( c = -4 \): \[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-4)}}{2(1)} = \frac{2 \pm \sqrt{4 + 16}}{2} = \frac{2 \pm \sqrt{20}}{2}. \] Simplify \(\sqrt{20}\): \[ \sqrt{20} = 2\sqrt{5}. \] Thus, \[ x = \frac{2 \pm 2\sqrt{5}}{2} = 1 \pm \sqrt{5}. \] So the quadratic factors into linear factors: \[ x - (1+\sqrt{5}) \quad \text{and} \quad x - (1-\sqrt{5}). \] **Step 3. Write the complete factorization:** Combining the factor \((x-1)\) from step 1, we have: \[ h(x) = (x-1)\left(x - (1+\sqrt{5})\right)\left(x - (1-\sqrt{5})\right). \] Thus, the factored form of \( h(x) \) is: \[ \boxed{(x-1)\left(x-(1+\sqrt{5})\right)\left(x-(1-\sqrt{5})\right)}. \]