Question

knowledge check - rational expressions
rewrite each expression below as as single, simplified fraction. state any excluded values.
a. \\(\frac{4}{2 - x} + \frac{x}{5}\\)
b. \\(\frac{x^2 - 6x + 8}{x^2 - 16} \div \frac{x^2 - 4x + 4}{x + 4}\\)
c. \\(\frac{\frac{2}{x} - \frac{3}{x^2}}{\frac{4}{x^3}}\\)

Explanation:

Part a

Step1: Find common denominator

Common denominator is $5(2-x)$

Step2: Rewrite fractions

$\frac{4 \cdot 5}{5(2-x)} + \frac{x(2-x)}{5(2-x)}$

Step3: Combine numerators

$\frac{20 + 2x - x^2}{5(2-x)}$

Step4: Identify excluded values

Denominator cannot be 0: $2-x
eq 0 \implies x
eq 2$

Step1: Rewrite division as multiplication

$\frac{x^2-6x+8}{x^2-16} \times \frac{x+4}{x^2-4x+4}$

Step2: Factor all polynomials

$\frac{(x-2)(x-4)}{(x-4)(x+4)} \times \frac{x+4}{(x-2)^2}$

Step3: Cancel common factors

Cancel $(x-4)$, $(x+4)$, $(x-2)$: $\frac{1}{x-2}$

Step4: Identify excluded values

Original denominators cannot be 0: $x^2-16
eq 0 \implies x
eq 4, -4$; $x^2-4x+4
eq 0 \implies x
eq 2$; divisor cannot be 0 so $x+4
eq 0$ (already covered)

Step1: Simplify numerator first

Find common denominator for numerator: $\frac{2x}{x^2} - \frac{3}{x^2} = \frac{2x-3}{x^2}$

Step2: Rewrite division as multiplication

$\frac{2x-3}{x^2} \times \frac{x^3}{4}$

Step3: Cancel common factors

Cancel $x^2$: $\frac{(2x-3)x}{4}$

Step4: Simplify expression

$\frac{2x^2 - 3x}{4}$

Step5: Identify excluded values

Original denominators cannot be 0: $x^2
eq 0$, $x^3
eq 0 \implies x
eq 0$

Answer:

Simplified fraction: $\frac{-x^2 + 2x + 20}{5(2-x)}$ (or $\frac{x^2 - 2x - 20}{5(x-2)}$ by factoring out -1 from numerator/denominator)
Excluded value: $x=2$

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Part b