Question

1. simplify the following rational expressions. be sure to completely reduce all fractions if possible.

a) \\(\frac{4x^2 - 2x}{(2x - 1)(2x + 1)}\\) b) \\(\frac{x^2 - 3x - 28}{(3x - 1)(x + 4)}\\)

c) \\(\frac{(x - 1)(x + 3)}{x^2 - 16} \cdot \frac{(x + 3)(x + 4)}{2x^2 + 3x - 5}\\) d) \\(\frac{x^2 - 4x}{(x + 2)(x - 1)} \div \frac{x - 4}{x + 2}\\)

e) \\(\frac{3x - 5}{x^2 - 25} + \frac{5}{x + 5}\\) f) \\(\frac{3}{x + 7} - \frac{2x + 17}{x^2 + 5x - 14}\\)

Explanation:

Part (a)

Step1: Factor the numerator

Factor out \(2x\) from \(4x^2 - 2x\), we get \(2x(2x - 1)\).
So the expression becomes \(\frac{2x(2x - 1)}{(2x - 1)(2x + 1)}\).

Step2: Cancel common factors

Cancel out the common factor \((2x - 1)\) from the numerator and the denominator.
We are left with \(\frac{2x}{2x + 1}\).

Step1: Factor the numerator

Factor \(x^2 - 3x - 28\). We need two numbers that multiply to \(-28\) and add to \(-3\). The numbers are \(-7\) and \(4\). So \(x^2 - 3x - 28=(x - 7)(x + 4)\).
The expression becomes \(\frac{(x - 7)(x + 4)}{(3x - 1)(x + 4)}\).

Step2: Cancel common factors

Cancel out the common factor \((x + 4)\) from the numerator and the denominator.
We are left with \(\frac{x - 7}{3x - 1}\).

Step1: Factor the denominators

Factor \(x^2 - 16\) as a difference of squares: \(x^2 - 16=(x - 4)(x + 4)\).
Factor \(2x^2 + 3x - 5\). We need two numbers that multiply to \(2\times(-5)=-10\) and add to \(3\). The numbers are \(5\) and \(-2\). So \(2x^2 + 3x - 5 = 2x^2+5x - 2x - 5=x(2x + 5)-1(2x + 5)=(2x + 5)(x - 1)\).
The expression becomes \(\frac{(x - 1)(x + 3)}{(x - 4)(x + 4)}\cdot\frac{(x + 3)(x + 4)}{(2x + 5)(x - 1)}\).

Step2: Cancel common factors

Cancel out \((x - 1)\) and \((x + 4)\) from the numerator and the denominator.
We are left with \(\frac{(x + 3)^2}{(x - 4)(2x + 5)}\) (after multiplying the remaining factors in the numerator: \((x + 3)\times(x + 3)=(x + 3)^2\) and in the denominator: \((x - 4)\times(2x + 5)\)).

Answer:

\(\frac{2x}{2x + 1}\)

Part (b)