Question

lesson 4 assignment continued
3 multiply or divide each. list any restrictions on the variables.
$\boldsymbol{a}$ $\frac{2x - 32}{x^2 - 10x + 24} cdot \frac{x^2 - 4x - 12}{10x + 20}$
$\boldsymbol{b}$ $\frac{6x}{x - 9} cdot \frac{x^2 - 10x + 9}{2x + 12} cdot \frac{x + 6}{x^4 + x^3 - 2x^2}$
$\boldsymbol{c}$ $\frac{8a^3b}{5c} div \frac{10ab^2}{3c}$
$\boldsymbol{d}$ $\frac{14x^2}{4x + 20} div \frac{7x^2 - 21x}{x^2 - 25}$
$\boldsymbol{e}$ $\frac{xy^2 - 6y^2}{2x + 18} div \frac{2xy + y}{8x + 24} div \frac{x^2 - 3x - 18}{2x^2 - 3x - 2}$

Explanation:

Part (a)

Step1: Factor all expressions

$\frac{2(x-16)}{(x-4)(x-6)} \cdot \frac{(x-6)(x+2)}{10(x+2)}$

Step2: Cancel common factors

Cancel $(x-6)$, $(x+2)$; simplify $\frac{2}{10}=\frac{1}{5}$
$\frac{(x-16)}{5(x-4)}$

Step3: Find variable restrictions

Denominators cannot be zero:
$x^2-10x+24=0 \implies x
eq4,6$;
$10x+20=0 \implies x
eq-2$;
Original numerators non-zero for cancellation: $x
eq16$ (from $2x-32=0$)

Step1: Factor all expressions

$\frac{6x}{x-9} \cdot \frac{(x-1)(x-9)}{2(x+6)} \cdot \frac{x+6}{x^2(x+2)(x-1)}$

Step2: Cancel common factors

Cancel $(x-9)$, $(x-1)$, $(x+6)$; simplify $\frac{6}{2}=3$
$\frac{3}{x(x+2)}$

Step3: Find variable restrictions

Denominators cannot be zero:
$x-9=0 \implies x
eq9$;
$2x+12=0 \implies x
eq-6$;
$x^4+x^3-2x^2=0 \implies x
eq0,-2,1$

Step1: Rewrite division as multiplication

$\frac{8a^3b}{5c} \cdot \frac{3c}{10ab^2}$

Step2: Multiply numerators/denominators

$\frac{8a^3b \cdot 3c}{5c \cdot 10ab^2}$

Step3: Cancel common factors

Cancel $a$, $b$, $c$; simplify $\frac{24}{50}=\frac{12}{25}$
$\frac{12a^2}{25b}$

Step4: Find variable restrictions

Denominators cannot be zero: $a
eq0$, $b
eq0$, $c
eq0$

Answer:

Simplified form: $\frac{x-16}{5(x-4)}$
Restrictions: $x
eq -2, 4, 6, 16$

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Part (b)