Problem 80 - Algebra for College Students

Question

Perform the indicated operations. $$\frac{9 w^{2}-64}{3 w^{2}-5 w-8} \cdot \frac{5 w^{2}+3 w-2}{25 w^{2}-4}$$

Step-by-Step Solution

Answer

\( \frac{(3w-8)(3w+8)(w+1)}{(3w+4)(w-2)(5w+2)} \)

Step 1 - Factorize the Numersator and Denominator of First Fraction

The first fraction given is \ \( \frac{9w^{2}-64}{3w^{2}-5w-8} \ \). Factorize the numerator and the denominator separately:- Numerator: Recognize the difference of squares: \ \( 9w^2-64 = (3w)^2 - 8^2 = (3w-8)(3w+8) \ \)- Denominator: Factorize by splitting the middle term: \ \( 3w^{2}-5w-8 = (3w+4)(w-2) \ \)

Step 2 - Factorize the Numerator and Denominator of Second Fraction

The second fraction given is \ \( \frac{5w^2 + 3w - 2}{25w^2 - 4} \ \). Factorize the numerator and the denominator separately:- Numerator: Factor by splitting the middle term: \ \( 5w^2 + 3w - 2 = (5w - 2)(w + 1) \ \)- Denominator: Recognize the difference of squares: \ \( 25w^2 - 4 = (5w)^2 - 2^2 = (5w-2)(5w+2) \ \)

Step 3 - Rewrite the Expression with Factored Forms

Substitute the factored forms into the original expression:\ \(\frac{(3w-8)(3w+8)}{(3w+4)(w-2)} \cdot \frac{(5w-2)(w+1)}{(5w-2)(5w+2)}\ \)

Step 4 - Cancel Common Factors

Identify and cancel common factors in the numerator and denominator:- Common factors: \ \( 5w-2 \ \)(Remove \ \( 5w-2 \ \) from both the numerator and denominator):\ \(\frac{(3w-8)(3w+8)}{(3w+4)(w-2)} \cdot \frac{(w+1)}{(5w+2)}\ \)

Step 5 - Multiply the Remaining Terms

Multiply the remaining terms in the numerator and the denominator: \ \( \frac{(3w-8)(3w+8)(w+1)}{(3w+4)(w-2)(5w+2)} \ \)

Key Concepts

Factoring Polynomials

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.

Difference of Squares

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.

Rational Expressions Multiplication

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.