Question

simplifying fractions: only factors can be cancelled!
$\frac{ax}{bx}=\frac{a}{b}$ $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}div\frac{d}{c}=\frac{a}{b}cdot\frac{d}{c}$ or use the lcd $\frac{\frac{a}{b}}{\frac{c}{d}}cdot bd=\frac{ad}{cb}$
only fractions with like denominators can be added or subtracted.
$\frac{a}{b}+\frac{c}{d}=\frac{ad}{bd}+\frac{cb}{db}=\frac{ad + cb}{bd}$
pitfalls:
$\frac{a + x}{bx}
eq\frac{a}{b}$ $\frac{a + x}{b + x}
eq\frac{a}{b}$
a number divided by 0 is a no! $\frac{n}{0}$, but 0 divided by a number is ok! $\frac{0}{k}$
examples:
a. $6m^{6}n + 7m^{5}n^{2}+2m^{4}n^{3}$ $m^{4}n(6m^{2}+7mn + 2n^{2})$
b. $\frac{-6(x^{2}-1)^{2}-(6x)(4x)(x^{2}-1)}{(x^{2}-1)^{4}}$

Explanation:

Step1: Simplify example A

Factor out the common factor $m^{4}n$ from $6m^{6}n + 7m^{5}n^{2}+2m^{4}n^{3}$. We get $m^{4}n(6m^{2}+7mn + 2n^{2})$.

Step2: Simplify example B

First, factor out $(x^{2}-1)$ from the numerator of $\frac{-6(x^{2}-1)^{2}-(6x)(4x)(x^{2}-1)}{(x^{2}-1)^{4}}$. The numerator becomes $(x^{2}-1)[-6(x^{2}-1)-24x^{2}]=(x^{2}-1)(-6x^{2}+6 - 24x^{2})=(x^{2}-1)(6 - 30x^{2})$. Then, cancel out the common factor $(x^{2}-1)$ between the numerator and the denominator. The expression simplifies to $\frac{6 - 30x^{2}}{(x^{2}-1)^{3}}$.

Answer:

Example A is factored as $m^{4}n(6m^{2}+7mn + 2n^{2})$. Example B simplifies to $\frac{6 - 30x^{2}}{(x^{2}-1)^{3}}$.