(b) Write \( x^{2}-7 x+3 \) as a product of linear factors. Write your answer in exact, simplified form. \( x^{2}-7 x+3=\square \)

To factor \( x^{2}-7x+3 \) into linear factors, we can use the quadratic formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -7 \), and \( c = 3 \). Plugging these values into the formula, we have: \[ x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \] \[ x = \frac{7 \pm \sqrt{49 - 12}}{2} \] \[ x = \frac{7 \pm \sqrt{37}}{2} \] Thus, the roots of the quadratic equation are \( \frac{7 + \sqrt{37}}{2} \) and \( \frac{7 - \sqrt{37}}{2} \). The factorization can be written as: \[ x^{2}-7x+3 = \left(x - \frac{7 + \sqrt{37}}{2}\right)\left(x - \frac{7 - \sqrt{37}}{2}\right) \]