For the polynomial below, -1 is a zero. \[ f(x)=x^{3}-5 x^{2}+x+7 \] Express \( f(x) \) as a product of linear factors. \( f(x)=\square \)

To express \( f(x) \) as a product of linear factors given that -1 is a zero, we can start by performing synthetic division of \( f(x) \) by \( x + 1 \). After the division, we get: 1. Set up synthetic division with -1: ``` -1 | 1 -5 1 7 | -1 6 -7 --------------------- 1 -6 7 0 ``` This results in a quotient of \( x^2 - 6x + 7 \). 2. Now, we need to factor \( x^2 - 6x + 7 \). We can find the roots using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = -6, c = 7 \): \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 28}}{2} = \frac{6 \pm \sqrt{8}}{2} = \frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2} \] Thus, the factors of \( f(x) \) can be expressed as: \[ f(x) = (x + 1)(x - (3 + \sqrt{2}))(x - (3 - \sqrt{2})) \] Putting everything together, we have: \[ f(x) = (x + 1)(x - (3 + \sqrt{2}))(x - (3 - \sqrt{2})) \]