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Factoring expressions is a crucial step in solving many algebraic problems, allowing you to break down complex expressions into simpler parts. In polynomial division, it often involves finding the greatest common factor (GCF) of the terms. Let's look at the given example: \[ -14 u^{3} v^{3} + 7 u^{2} v^{3} + 21 u v + 56 \]Here, we first identify common factors in each term. In this case, each term includes a '7', which can be factored out. Additionally, the terms involving variables share 'u' and 'v' to certain degrees. By extracting these common terms, we simplify the expression initially: \[ 7uv(-2u^{2}v^{2} + u v^{2} + 3 + 8) \]Factoring can make other processes straightforward, like simplifying and cancelling, as you're often dealing with smaller numbers and expressions.

Simplifying rational expressions means rewriting them in a less complicated form, while retaining their original value. A rational expression is a fraction where the numerator and the denominator are polynomials, just like in our original problem:\[ \frac{-14 u^{3} v^{3} + 7 u^{2} v^{3} + 21 u v + 56}{7 u^{2} v} \]First, ensure the expression is factorized, as we did. Then, check for any common factors both in the numerator and the denominator that could be canceled out. Once factored, simplifying the fraction can involve reducing each term independently or collectively, so the overall expression becomes easy to work with. In our exercise, this reduction helped transform the expression into:\[ -2u v^{2} + v^{2} + \frac{3}{u} + \frac{8}{u} \]Simplification makes it easier to identify the most important characteristics of an expression and paves the way for further mathematical analysis or solution processes.

Canceling common factors in a fraction is an important step in simplifying expressions because it reduces the fraction to its simplest form. It requires carefully examining the factored terms and eliminating those that appear both in the numerator and the denominator.From the expression we began with:\[ \frac{7uv(-2u^{2}v^{2} + u v^{2} + 3 + 8)}{7 u^{2} v} \]Notice '7uv' in the numerator and a similar term '7u^2v' in the denominator. Here, you can cancel the '7uv' from the top and bottom, considering its squared power and additional factor, leaving:\[ \frac{1}{u}(-2u^{2}v^{2} + u v^{2} + 3 + 8) \]This process reduces the complexity of a complex fraction and makes it clearer to identify and work with remaining terms. Reducing fractions by canceling is particularly important in algebra to ensure subsequent calculations are straightforward.