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Factoring polynomials is akin to breaking down a number into its prime components, but in algebraic terms. The goal is to express a polynomial as a product of simpler polynomials. For instance, consider the polynomial \( 3w^2 - 5w - 8 \). To factor this, we look for two binomials whose product gives the original polynomial.
First, identify two numbers that multiply to \( ac \) (where \( a \) is the coefficient of \( w^2 \) and \( c \) is the constant term) and add to \( b \) (the coefficient of \( w \)).

• Here, \( ac = 3 \, -8 = -24 \) and \( b = -5 \).
• The numbers \( -8 \) and \( 3 \) fit since \( -8 * 3 = -24 \) and \( -8 + 3 = -5 \).

Next, rewrite the middle term using these numbers: \( 3w^2 - 8w + 3w - 8 \). Then, factor by grouping: \( (3w^2 - 8w) + (3w - 8) = (w-2)(3w+4) \).
This method breaks down complex polynomials into more manageable factors.

The difference of squares is a special factoring rule used when a polynomial is in the form \( a^2 - b^2 \). The formula is \(a^2 - b^2 = (a - b)(a + b) \).
Let's take \( 9w^2 - 64 \) from the original exercise. Notice that both terms, \( 9w^2 \) and \( 64 \), are perfect squares:

• \( 9w^2 \) is \( (3w)^2 \)
• \( 64 \) is \( 8^2 \)

Therefore, we can apply the difference of squares formula: \( 9w^2 - 64 = (3w)^2 - 8^2 = (3w - 8)(3w + 8) \). This method is quick and useful for simplifying expressions with recognizable patterns.

Multiplying rational expressions involves multiplying the numerators together and the denominators together. However, simplifying before multiplying can save significant effort.
Let's revisit the given problem:\( \frac{(3w-8)(3w+8)}{(3w+4)(w-2)} \cdot \frac{(5w-2)(w+1)}{(5w-2)(5w+2)} \).
Before multiplying, eliminate any common factors present in both the numerators and denominators.

• Here, \( 5w - 2 \) is a common factor
• After canceling out, we get: \( \frac{(3w-8)(3w+8)}{(3w+4)(w-2)} \cdot \frac{(w+1)}{(5w+2)} \).

Finally, multiply the remaining terms: \( \frac{(3w-8)(3w+8)(w+1)}{(3w+4)(w-2)(5w+2)} \). Simplifying before multiplying keeps the expression manageable and reduces potential errors.