91Ó°ÊÓ

When simplifying algebraic fractions, one crucial step is factoring polynomials. Factoring helps us break down complex expressions into simpler components.
For example, we have the expressions $$ 5b + 15 $$ and $$ 4b + 4 $$ in our initial fraction. To factor these:
- For $$ 5b + 15 $$, we can factor out the common factor 5, giving us $$ 5(b + 3) $$.
- For $$ 4b + 4 $$, we factor out 4, giving us $$ 4(b + 1) $$.
Similarly, the second fraction, $$ 2b + 6 $$ and $$ 7b + 7 $$, can be factored as:
- $$ 2b + 6 $$ becomes $$ 2(b + 3) $$
- $$ 7b + 7 $$ becomes $$ 7(b + 1) $$
Factoring transforms our problem into simpler forms, making it easier to see common factors and cancel them out.

When dividing fractions, we use the reciprocal of the second fraction, turning the division into multiplication.
The reciprocal of a fraction means swapping its numerator and denominator.
For instance, the reciprocal of \ \frac{2b+6}{7b+7} \ is \ \frac{7b+7}{2b+6} \.
This transformation lets us rewrite the expression \ \(\frac{5b+ 15}{4b+ 4} \div \frac{2b+ 6}{7b+ 7}\) \ as multiplication: \ \(\frac{5b + 15}{4b + 4} \times \frac{7b + 7}{2b + 6}\). \

After factoring and rewriting the division as multiplication, the next step is canceling common factors. This simplifies the expression more by removing identical factors in the numerator and denominator.
From our factored expression \ \(\frac{5(b + 3)}{4(b + 1)} \times \frac{7(b + 1)}{2(b + 3)}\), \ we notice:
- \ \((b+3)\) \ appears in both numerator and denominator.
- \ \((b+1)\) \ appears in both numerator and denominator.
These common factors can be canceled, leaving us with: \ \(\frac{5}{4} \times \frac{7}{2}\). \ Multiplying the remaining fractions gives \ \(\frac{35}{8}\). \ Canceling common factors helps reduce the complexity and facilitates easier calculations.