Let's work through the problems one by one. Here's the breakdown of each problem with its solution:

1. Given $f(x) = 3x^2 + 7x$ and $g(x) = 2x^2 - x - 1$, find $(f + g)(x)$

We add the two functions: $(f + g)(x) = f(x) + g(x) = (3x^2 + 7x) + (2x^2 - x - 1)$ Simplify: $(f + g)(x) = (3x^2 + 2x^2) + (7x - x) - 1 = 5x^2 + 6x - 1$ So, $(f + g)(x) = 5x^2 + 6x - 1$.

2. Given $g(x) = 6x + 4$ and $f(x) = 2x - 7$, find $(fg)(x)$

We multiply the two functions: $(fg)(x) = f(x) \cdot g(x) = (2x - 7)(6x + 4)$ Use the distributive property: $(fg)(x) = 2x(6x + 4) - 7(6x + 4) = 12x^2 + 8x - 42x - 28$ Simplify: $(fg)(x) = 12x^2 - 34x - 28$ So, $(fg)(x) = 12x^2 - 34x - 28$.

3. Given $g(x) = 6x + 4$ and $f(x) = 2x - 7$, find $g(x) - f(x)$

Subtract the two functions: $(g - f)(x) = g(x) - f(x) = (6x + 4) - (2x - 7)$ Distribute the negative sign: $(g - f)(x) = 6x + 4 - 2x + 7 = 4x + 11$ So, $g(x) - f(x) = 4x + 11$.

4. Given $f(x) = x^2 - 2x + 1$ and $g(x) = x - 1$, find $(f/g)(x)$

We divide the two functions: $\frac{f(x)}{g(x)} = \frac{x^2 - 2x + 1}{x - 1}$ Factor the numerator: $\frac{(x - 1)^2}{x - 1}$ Cancel out the $x - 1$ terms: $\frac{f(x)}{g(x)} = x - 1$ So, $\frac{f(x)}{g(x)} = x - 1$ (for $x \neq 1$).

5. Given $f(x) = x + 4$ and $g(x) = x - 4$, find $(fg)(x)$

Multiply the two functions: $(fg)(x) = f(x) \cdot g(x) = (x + 4)(x - 4)$ This is a difference of squares: $(fg)(x) = x^2 - 16$ So, $(fg)(x) = x^2 - 16$.

Let me know if you need more details on any of these steps or if you'd like me to continue with the remaining problems.

Questions:

1. How would you factor quadratic expressions to simplify division?
2. How can you identify when to use the difference of squares technique in multiplication?
3. What are the common steps to multiply polynomials?
4. When simplifying rational expressions, when should terms be canceled?
5. How does the distributive property help in expanding polynomials?

Tip: Always check for factoring opportunities before simplifying polynomial divisions!