Problem 95 - Algebra for College Students

Question

Perform the indicated operations. $$ \frac{4 x^{2}+9}{4 x^{2}-9} \cdot \frac{4 x^{2}+12 x+9}{2 x^{2}+3 x} $$

Step-by-Step Solution

Answer

\(\frac{(4x^2 + 9)(2x + 3)}{(2x - 3)x}\)

Step 1: Factor common terms

Factor each polynomial in the expression.\[4x^2 + 9 = (2x)^2 + 3^2\] is a sum of squares and cannot be factored further.\[4x^2 - 9 = (2x)^2 - 3^2 = (2x - 3)(2x + 3)\] by the difference of squares formula \[4x^2 + 12x + 9 = (2x + 3)^2\] \[2x^2 + 3x = x(2x + 3)\].

Step 2: Rewrite the Expression with Factored Forms

Substitute the factored forms into the original expression: \[\frac{4 x^2 + 9}{4 x^2 - 9} \times \frac{4 x^2 + 12 x + 9}{2 x^2 + 3 x} = \frac{4 x^2 + 9}{(2x - 3)(2x + 3)} \times \frac{(2x + 3)^2}{x (2x + 3)}\].

Step 3: Simplify the Expression

Cancel out the common factors in the numerator and the denominator. First, cancel \(2x + 3\) from \((2x + 3)^2\) and\(2x + 3\) remaining in the denominator: \[= \frac{4 x^2 + 9}{(2x - 3)} \times \frac{2x + 3}{x}\].

Step 4: Multiply the Remaining Factors

Multiply the numerators and the denominators: \[= \frac{(4x^2 + 9)(2x + 3)}{(2x - 3)x}\].

Key Concepts

polynomial factorization

Polynomial factorization is a technique where we express a polynomial as a product of its factors. This means breaking down a complex polynomial into simpler ones whose product equals the original
. It's like finding the building blocks of the polynomial, making it easier to simplify or solve.

For example, in our exercise, \(4x^2 - 9\) is factored as \((2x - 3)(2x + 3)\) using the difference of squares formula. Factorizing helps us see common terms and cancel them out for simplification.

difference of squares

The difference of squares is a special factoring technique used when we have a polynomial in the form \(a^2 - b^2\). This can be rewritten as \( (a - b)(a + b) \). This method is powerful for simplifying polynomials quickly.

In our given exercise, we see the expression \(4x^2 - 9\). Here, \(a = 2x\) and \(b = 3\). Applying the difference of squares formula, we factorize it as \((2x - 3)(2x + 3)\). This reveals pairs of factors that are easier to handle in multiplication or division.

simplifying fractions

Simplifying fractions involves reducing a fraction to its simplest form. This means making the numerator and the denominator as small as possible while keeping their ratio the same.

In the context of polynomials, this involves factoring both the numerator and the denominator, then canceling out any common factors.

For instance, in our exercise, we cancel out \(2x + 3\) terms from the factored forms: \(\frac{4 x^2 + 9}{(2x - 3)(2x + 3)} \times \frac{(2x + 3)^2}{x (2x + 3)} = \frac{4 x^2 + 9}{(2x - 3)} \times \frac{2x + 3}{x}\). This process of canceling and simplifying makes the problem much easier to solve. By carefully factoring and canceling, we get a simplified version that is much easier to work with.