Problem 80 - Elementary Algebra

Question

For exercises 39-82, simplify. $$ \frac{16 x^{2}+8 x+1}{8 x^{2}-10 x-3} \div \frac{4 x^{2}+17 x+4}{2 x^{2}+x-6} $$

Step-by-Step Solution

Answer

Simplify to \(\frac{(4x+1)(x+2)}{x+4}\).

Step 1: Factorize the Numerators

First, factorize the numerators of both fractions. For the numerator of the first fraction, we have: \(16x^2 + 8x + 1 = (4x + 1)^2\) For the numerator of the second fraction, we have: \(4x^2 + 17x + 4\). To factorize: Find two numbers that multiply to 16 and add to 17. The numbers are 16 and 1. So we get: \(4x^2 + 16x + x + 4 = (4x+1)(x+4)\)

Step 2: Factorize the Denominators

Next, factorize the denominators of both fractions. For the denominator of the first fraction, we have: \(8x^2 - 10x - 3\). To factorize: Find two numbers that multiply to -24 and add to -10. The numbers are -12 and 2. So we get: \(8x^2 - 12x + 2x - 3 = 4x(2x-3) + 1(2x-3) = (4x + 1)(2x - 3)\) For the denominator of the second fraction, we have:\(2x^2 + x - 6\). To factorize: Find two numbers that multiply to -12 and add to 1. The numbers are 4 and -3. So we get: \(2x^2 + 4x - 3x - 6 = 2x(x+2) - 3(x+2) = (2x - 3)(x + 2)\)

Step 3: Rewrite the Division as Multiplication

When dividing fractions, multiply by the reciprocal of the divisor. So rewrite the expression as:\(\frac{(4x + 1)^2}{(4x + 1)(2x - 3)} \times \frac{(2x - 3)(x+2)}{(4x+1)(x+4)}\)

Step 4: Simplify

Now, cancel common factors in the numerator and denominator: \(\frac{(4x + 1)^2}{(4x + 1)(2x - 3)} \times \frac{(2x - 3)(x+2)}{(4x + 1)(x+4)}\)The \( (4x + 1) \) in the numerator and denominator cancel out, leaving: \(\frac{4x + 1}{2x - 3} \times \frac{(2x - 3)(x+2)}{x+4}\)The \(2x - 3\) in the numerator and denominator cancel out, leaving:\(\frac{(4x + 1)(x+2)}{x+4}\)This is the simplified form.

Key Concepts

Factorization

Factorization is a key concept in algebra that involves breaking down a complex expression into simpler components, or factors, that when multiplied together give the original expression.
When simplifying rational expressions, the first step is often to factorize both the numerators and the denominators.
In our exercise, we started by factorizing the numerators of both fractions: For the first fraction, the numerator is: {n}