Let's solve each of these questions step by step.

Question 1: Find $f(x) + g(x)$

Given:

• $f(x) = 2x - 7$
• $g(x) = x^2 + 2x - 3$

To find $f(x) + g(x)$, simply add the two expressions together:

$f(x) + g(x) = (2x - 7) + (x^2 + 2x - 3)$ Combine like terms:

$= x^2 + 2x + 2x - 7 - 3$ $= x^2 + 4x - 10$

So, the answer is:

b. $x^2 + 4x - 10$

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Question 2: Find $g(f(x))$

Given:

• $f(x) = x - 3$
• $g(x) = x^3$

We are asked to find $g(f(x))$, which means substituting $f(x) = x - 3$ into $g(x)$. So, we compute $g(f(x))$ as:

$g(f(x)) = g(x - 3)$ Since $g(x) = x^3$, substitute $x - 3$ into this:

$g(x - 3) = (x - 3)^3$

Now expand $(x - 3)^3$:

$(x - 3)^3 = (x - 3)(x - 3)(x - 3)$ First, expand $(x - 3)(x - 3)$:

$(x - 3)(x - 3) = x^2 - 6x + 9$

Now multiply this result by $(x - 3)$:

$(x^2 - 6x + 9)(x - 3)$ Distribute each term:

$= x^3 - 3x^2 - 6x^2 + 18x + 9x - 27$ $= x^3 - 9x^2 + 27x - 27$

So the answer is:

d. $x^3 - 9x^2 + 27x - 27$

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Question 3: Simplify $\frac{2 - 4i}{1 + 3i}$

To simplify the expression $\frac{2 - 4i}{1 + 3i}$, multiply both the numerator and the denominator by the conjugate of the denominator, $1 - 3i$:

$\frac{2 - 4i}{1 + 3i} \times \frac{1 - 3i}{1 - 3i}$

This gives:

$\frac{(2 - 4i)(1 - 3i)}{(1 + 3i)(1 - 3i)}$

First, simplify the denominator:

$(1 + 3i)(1 - 3i) = 1^2 - (3i)^2 = 1 - (-9) = 1 + 9 = 10$

Now, simplify the numerator by expanding $(2 - 4i)(1 - 3i)$:

$(2 - 4i)(1 - 3i) = 2(1) - 2(3i) - 4i(1) + 4i(3i)$ $= 2 - 6i - 4i + 12i^2$ Since $i^2 = -1$, we have:

$= 2 - 6i - 4i - 12$ $= -10 - 10i$

So, the expression becomes:

$\frac{-10 - 10i}{10} = -1 - i$

Thus, the answer is:

c. $-1 - i$

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Summary of Answers:

1. b. $x^2 + 4x - 10$
2. d. $x^3 - 9x^2 + 27x - 27$
3. c. $-1 - i$

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Would you like more details on any of the solutions? Here are some related questions to explore:

1. How do you solve for $f(x) + g(x)$ when $f(x)$ and $g(x)$ are more complicated polynomials?
2. What is the general formula for $g(f(x))$ when $f(x)$ is a linear function?
3. How do you expand the cube of a binomial like $(x - 3)^3$?
4. What is the role of the conjugate in simplifying complex fractions?
5. Can you solve other types of expressions involving complex numbers, such as $\frac{a + bi}{c + di}$?

Tip: When simplifying expressions with complex numbers, always use the conjugate of the denominator to eliminate imaginary parts from the denominator.