Let's solve and simplify the given expression:

$\frac{6x - 3}{4x^2 - 1} + \frac{x^2 - 49}{(2x^2 + x)(3x - 21)}$

Step 1: Factorize the denominators

• $4x^2 - 1$ can be factorized as a difference of squares: $(2x - 1)(2x + 1)$.
• $x^2 - 49$ is also a difference of squares: $(x - 7)(x + 7)$.
• $(2x^2 + x)(3x - 21)$ can be further factorized:
  • Factor out $x$ from $2x^2 + x$ to get $x(2x + 1)$.
  • Factor out $3$ from $3x - 21$ to get $3(x - 7)$.

Thus, $(2x^2 + x)(3x - 21) = x(2x + 1) \cdot 3(x - 7) = 3x(x - 7)(2x + 1)$.

Step 2: Find a common denominator The least common denominator (LCD) for both fractions is: $(2x - 1)(2x + 1) \cdot 3x(x - 7)$

Step 3: Rewrite the fractions over the common denominator $\frac{(6x - 3)(3x(x - 7))}{LCD} + \frac{(x - 7)(x + 7)}{(2x - 1)(2x + 1)} \cdot \frac{3x}{3x}$

Step 4: Expand and simplify the numerators

• $(6x - 3)(3x(x - 7)) = 3(2x - 1) \cdot 3x(x - 7) = 9x(2x - 1)(x - 7)$
• $(x - 7)(x + 7) \cdot 3x = 3x(x^2 - 49) = 3x(x - 7)(x + 7)$

The addition then becomes: $\frac{9x(2x - 1)(x - 7) + 3x(x - 7)(x + 7)}{(2x - 1)(2x + 1) \cdot 3x(x - 7)}$

Step 5: Combine and simplify further $\frac{9x(2x - 1)(x - 7) + 3x(x - 7)(x + 7)}{LCD}$ Simplify terms with $(x - 7)$ and other common factors: $\frac{3x(x - 7)(3(2x - 1) + (x + 7))}{LCD}$ $\frac{3x(x - 7)(6x - 3 + x + 7)}{LCD}$ $\frac{3x(x - 7)(7x + 4)}{LCD}$

The final simplified expression: $\frac{3x(x - 7)(7x + 4)}{(2x - 1)(2x + 1) \cdot 3x(x - 7)}$

Further reduction by cancelling out $3x$ and $(x - 7)$: $\frac{7x + 4}{(2x - 1)(2x + 1)}$

Would you like to delve into any steps in more detail or have any questions about this solution?

Here are some questions that might further deepen your understanding:

1. How do you factorize a cubic polynomial?
2. What is the significance of finding the least common denominator in fraction addition?
3. Can every polynomial be factorized into linear factors?
4. How can we verify the simplified form is correct?
5. Why is it important to cancel out common factors in a fraction?

Tip: Always check for the greatest common factor first when simplifying expressions; it can often make the algebraic manipulation significantly easier.