Question

1. simplify(reduce) the expression. list all domain restrictions. \\(\frac{5x}{x + 4} \cdot \frac{3x^2 + 12x}{7x - 7} \div \frac{3}{x^2 - 2x + 1}\\)\

restriction(s): (select all that apply)\
\\(\square x \
eq 1\\)\
\\(\square x \
eq -\frac{1}{2}\\)\
\\(\square\\) no restrictions\
\\(\square x \
eq -1\\)\
\\(\square x \
eq 0\\)\
\\(\square x \
eq -4\\)

Explanation:

Step1: Rewrite division as multiplication

$\frac{5x}{x+4} \cdot \frac{3x^2+12x}{7x-7} \cdot \frac{x^2-2x+1}{3}$

Step2: Factor all polynomials

$\frac{5x}{x+4} \cdot \frac{3x(x+4)}{7(x-1)} \cdot \frac{(x-1)^2}{3}$

Step3: Cancel common factors

$\frac{5x}{\cancel{x+4}} \cdot \frac{\cancel{3}x\cancel{(x+4)}}{7\cancel{(x-1)}} \cdot \frac{(x-1)^{\cancel{2}}}{\cancel{3}}$
Simplify remaining terms: $5x \cdot \frac{x}{7} \cdot (x-1)$

Step4: Multiply remaining terms

$\frac{5x \cdot x \cdot (x-1)}{7} = \frac{5x^2(x-1)}{7}$

Step5: Find domain restrictions

Denominators cannot be zero:

1. $x+4

eq 0 \implies x
eq -4$

1. $7x-7

eq 0 \implies x
eq 1$

1. $x^2-2x+1

eq 0 \implies (x-1)^2
eq 0 \implies x
eq 1$

1. Original divisor $\frac{3}{x^2-2x+1}$ cannot be zero (numerator 3≠0, no new restriction)

Also, factors canceled from denominators: $3x(x+4)
eq 0 \implies x
eq 0, x
eq -4$

Answer:

Simplified expression: $\frac{5x^2(x-1)}{7}$
Domain restrictions: $x
eq -4$, $x
eq 0$, $x
eq 1$