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}=\frac{5x^2(x - 1)}{7}\\) restriction(s): (select all that apply) \\(\square x\

eq1\\) \\(\square x\
eq-\frac{1}{2}\\) \\(\square\\) no restrictions \\(\square x\
eq-1\\) \\(\square x\
eq0\\) \\(\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

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

Step4: Multiply remaining terms

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

Step5: Identify domain restrictions

Denominators cannot be 0:

1. $x+4

eq 0 \implies x
eq -4$

1. $7x-7

eq 0 \implies x
eq 1$

1. $3

eq 0$ (no restriction)

1. Original divisor $\frac{3}{x^2-2x+1}

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

1. Numerators of original fractions cannot make denominators 0, and $5x$ and $3x^2+12x$ have $x=0$ as a root, but $x=0$ makes original expression defined? No: $x=0$ gives $\frac{0}{4} \cdot \frac{0}{-7} \div \frac{3}{1}=0$, which is valid. $7x-7=0$ at $x=1$, $x+4=0$ at $x=-4$, $x^2-2x+1=0$ at $x=1$. Also, $7x-7=7(x-1)$, so $x

eq 1$; $x+4
eq 0 \implies x
eq -4$; also, from $3x^2+12x=3x(x+4)$, no new restriction, and $x=0$ is allowed. Wait, also check if any canceled factors had restrictions: we canceled $(x+4)$, $(x-1)$, $3$. So all restrictions are $x
eq -4$, $x
eq 1$.

Answer:

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