Given the following functions, find each: \( f(x)=x^{2}+3 x-28 \) \( g(x)=x+7 \) \( (f+g)(x)=x^{2}+4 x-21 \) \( (f-g)(x)=x^{2}+2 x-35 \) \( (f \cdot g)(x)=x^{3}+10 x^{2}-7 x-196 \) \( \left(\frac{f}{g}\right)(x)= \)

To find \( \left(\frac{f}{g}\right)(x) \), we start by substituting the expressions for \( f(x) \) and \( g(x) \): \[ f(x) = x^2 + 3x - 28 \] \[ g(x) = x + 7 \] Now, we can express \( \left(\frac{f}{g}\right)(x) \) as: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2 + 3x - 28}{x + 7} \] Next, we can simplify this fraction by performing polynomial long division or factoring the numerator if possible. Factoring \( f(x) \): To factor \( x^2 + 3x - 28 \), we look for two numbers that multiply to \(-28\) and add to \(3\). Those numbers are \(7\) and \(-4\). So, we can write: \[ x^2 + 3x - 28 = (x + 7)(x - 4) \] This allows us to rewrite the expression as: \[ \frac{(x + 7)(x - 4)}{x + 7} \] Now, we can cancel \( x + 7 \) from the numerator and the denominator (as long as \( x \neq -7 \)): \[ \left(\frac{f}{g}\right)(x) = x - 4 \quad (x \neq -7) \] Thus, the final result is: \[ \left(\frac{f}{g}\right)(x) = x - 4 \quad \text{for } x \neq -7 \]