The problem asks to simplify the given rational expressions. We are given two expressions to simplify. (e) $\frac{2x^2 + 17x + 21}{(x+2)(x^2-9)}$ (i) $\frac{3x - 10x - 24}{x(x^2-4)}$

1. Problem Description

The problem asks to simplify the given rational expressions. We are given two expressions to simplify.

(e) $\frac{2x^2 + 17x + 21}{(x+2)(x^2-9)}$

(i) $\frac{3x - 10x - 24}{x(x^2-4)}$

2. Solution Steps

(e)

First, we factor the numerator $2x^2 + 17x + 21$.

We are looking for two numbers that multiply to $2 \cdot 21 = 42$ and add up to $17$. These numbers are $3$ and $14$.

So, $2x^2 + 17x + 21 = 2x^2 + 3x + 14x + 21 = x(2x+3) + 7(2x+3) = (x+7)(2x+3)$.

Next, we factor the denominator $(x+2)(x^2-9)$.

Since $x^2 - 9$ is a difference of squares, $x^2 - 9 = (x-3)(x+3)$.

So the denominator is $(x+2)(x-3)(x+3)$.

Then the expression is

$\frac{(x+7)(2x+3)}{(x+2)(x-3)(x+3)}$.

There are no common factors to cancel, so the simplified form is $\frac{(x+7)(2x+3)}{(x+2)(x-3)(x+3)}$.

(i)

First, we simplify the numerator $3x - 10x - 24 = -7x - 24$.

Then, we factor the denominator $x(x^2-4)$.

Since $x^2-4$ is a difference of squares, $x^2-4 = (x-2)(x+2)$.

So the denominator is $x(x-2)(x+2)$.

The expression becomes $\frac{-7x-24}{x(x-2)(x+2)}$.

There are no common factors to cancel. So the simplified form is $\frac{-7x-24}{x(x-2)(x+2)}$.

3. Final Answer

(e) $\frac{(x+7)(2x+3)}{(x+2)(x-3)(x+3)}$

(i) $\frac{-7x-24}{x(x-2)(x+2)}$