The problem asks us to simplify rational expressions. We are given several fractions involving polynomials and need to simplify them by factoring and cancelling common factors. The expressions are: (e) $\frac{2x^2 + 17x + 21}{(x+2)(x^2-9)}$ (f) $\frac{7}{(2x-3)(x+2)}$ (g) $\frac{2x-3}{x^2-1}$ (h) $\frac{37x-81}{(2x-3)(x^2+4x-21)}$ (i) $\frac{3x^2 - 10x - 24}{x(x^2-4)}$ (j) $\frac{2x^2 - 9x - 35}{x^3 + 2x^2 - 5x - 6}$

Okay, I will solve the problems as requested.

1. Problem Description

The problem asks us to simplify rational expressions. We are given several fractions involving polynomials and need to simplify them by factoring and cancelling common factors. The expressions are:

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

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

(g) $\frac{2x-3}{x^2-1}$

(h) $\frac{37x-81}{(2x-3)(x^2+4x-21)}$

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

(j) $\frac{2x^2 - 9x - 35}{x^3 + 2x^2 - 5x - 6}$

2. Solution Steps

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

First, we factor the numerator and denominator.

$2x^2 + 17x + 21 = (2x+3)(x+7)$

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

Thus, $\frac{2x^2 + 17x + 21}{(x+2)(x^2-9)} = \frac{(2x+3)(x+7)}{(x+2)(x+3)(x-3)}$

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

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

The expression is already simplified.

(g) $\frac{2x-3}{x^2-1}$

We factor the denominator as $x^2-1 = (x-1)(x+1)$.

Thus, $\frac{2x-3}{x^2-1} = \frac{2x-3}{(x-1)(x+1)}$.

There are no common factors to cancel.

(h) $\frac{37x-81}{(2x-3)(x^2+4x-21)}$

We factor the quadratic in the denominator.

$x^2+4x-21 = (x+7)(x-3)$

Thus, $\frac{37x-81}{(2x-3)(x^2+4x-21)} = \frac{37x-81}{(2x-3)(x+7)(x-3)}$

There are no obvious common factors to cancel.

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

We factor the numerator and denominator.

$3x^2 - 10x - 24 = (3x+8)(x-3)$

$x^2-4 = (x-2)(x+2)$

Thus, $\frac{3x^2 - 10x - 24}{x(x^2-4)} = \frac{(3x+8)(x-3)}{x(x-2)(x+2)}$

There are no common factors to cancel.

(j) $\frac{2x^2 - 9x - 35}{x^3 + 2x^2 - 5x - 6}$

We factor the numerator.

$2x^2 - 9x - 35 = (2x+5)(x-7)$

We look for roots of the denominator by using the Rational Root Theorem. Try $x=2$: $2^3 + 2(2^2) - 5(2) - 6 = 8 + 8 - 10 - 6 = 0$.

So, $x-2$ is a factor. We divide $x^3 + 2x^2 - 5x - 6$ by $x-2$.

$x^3 + 2x^2 - 5x - 6 = (x-2)(x^2 + 4x + 3) = (x-2)(x+1)(x+3)$

Thus, $\frac{2x^2 - 9x - 35}{x^3 + 2x^2 - 5x - 6} = \frac{(2x+5)(x-7)}{(x-2)(x+1)(x+3)}$

There are no common factors to cancel.

3. Final Answer

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

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

(g) $\frac{2x-3}{(x-1)(x+1)}$

(h) $\frac{37x-81}{(2x-3)(x+7)(x-3)}$

(i) $\frac{(3x+8)(x-3)}{x(x-2)(x+2)}$

(j) $\frac{(2x+5)(x-7)}{(x-2)(x+1)(x+3)}$