17. what are the restrictions on x for the following expression? select two answers.
$\frac{x^2 + 8x + 15}{x^2 - 4x}$
a. $x \
eq -5$ b. $x \
eq -3$ c. $x \
eq 0$
d. $x \
eq 4$ e. $x \
eq -8$ f. $x \
eq -4$

simplify each rational expression. identify any restrictions on x.
18. $\frac{x^2 + 13x + 36}{x^2 + 23 - 63}$ 19. $\frac{5x - 15}{x^2 - x - 6}$

find each product or quotient. identify any restrictions on x.
20. $\frac{x^2 - 3x - 10}{2x - 6} \cdot \frac{8x + 10}{x^2 - 5x + 6}$ 21. $\frac{x(x - 2)}{x^2 + 7x - 18} \cdot \frac{x^2 - 81}{3(x - 9)(x - 2)}$ 22. $\frac{x^2 + 10x + 21}{x^2 - 68 - 16} \div \frac{x + 3}{x^2 - 64}$

Question

17. what are the restrictions on x for the following expression? select two answers.
$\frac{x^2 + 8x + 15}{x^2 - 4x}$
a. $x \
eq -5$  b. $x \
eq -3$  c. $x \
eq 0$
d. $x \
eq 4$  e. $x \
eq -8$  f. $x \
eq -4$

simplify each rational expression. identify any restrictions on x.
18. $\frac{x^2 + 13x + 36}{x^2 + 23 - 63}$  19. $\frac{5x - 15}{x^2 - x - 6}$

find each product or quotient. identify any restrictions on x.
20. $\frac{x^2 - 3x - 10}{2x - 6} \cdot \frac{8x + 10}{x^2 - 5x + 6}$  21. $\frac{x(x - 2)}{x^2 + 7x - 18} \cdot \frac{x^2 - 81}{3(x - 9)(x - 2)}$  22. $\frac{x^2 + 10x + 21}{x^2 - 68 - 16} \div \frac{x + 3}{x^2 - 64}$

Accepted Answer

### Question 17
# Explanation:
## Step 1: Factor numerator and denominator
Factor $x^2 + 8x + 15$: $x^2 + 8x + 15=(x + 3)(x + 5)$
Factor $x^2 - 4x$: $x^2 - 4x=x(x - 4)$
## Step 2: Find restrictions (denominator ≠ 0)
For $x(x - 4)
eq0$, we have $x
eq0$ (from $x = 0$) and $x
eq4$ (from $x - 4=0$). Also, the numerator's factors don't affect restrictions here, but we check if any of the numerator's roots make denominator zero (they don't here). So restrictions are $x
eq0$ and $x
eq4$, and also, wait, no—wait, the original expression is $\frac{x^2 + 8x + 15}{x^2 - 4x}$, so denominator is $x(x - 4)$, so $x
eq0$ and $x
eq4$. But also, when simplifying, but the question is about restrictions on $x$ for the expression, so denominator can't be zero. So the restrictions are $x
eq0$ (option C) and $x
eq4$ (option D), and also, wait, the numerator is $(x + 3)(x + 5)$, so no conflict. Wait, but the options: A. $x
eq - 5$, B. $x
eq - 3$, C. $x
eq0$, D. $x
eq4$, E. $x
eq - 8$, F. $x
eq - 4$. So the denominator is $x(x - 4)$, so $x
eq0$ (C) and $x
eq4$ (D). So the two answers are C and D.
# Answer: C. $x
eq 0$, D. $x
eq 4$

### Question 18
# Explanation:
## Step 1: Factor numerator and denominator
Factor $x^2 + 13x + 36$: $x^2 + 13x + 36=(x + 4)(x + 9)$
Factor $x^2 + 2x - 63$: $x^2 + 2x - 63=(x + 9)(x - 7)$
## Step 2: Simplify the expression
$\frac{(x + 4)(x + 9)}{(x + 9)(x - 7)}=\frac{x + 4}{x - 7}$ (cancel $x + 9$, $x
eq - 9$)
## Step 3: Find restrictions (denominator ≠ 0)
Original denominator: $(x + 9)(x - 7)
eq0$ ⇒ $x
eq - 9$ and $x
eq7$
# Answer: Simplified: $\frac{x + 4}{x - 7}$, Restrictions: $x
eq - 9$, $x
eq7$

### Question 19
# Explanation:
## Step 1: Factor numerator and denominator
Factor $5x - 15$: $5x - 15 = 5(x - 3)$
Factor $x^2 - x - 6$: $x^2 - x - 6=(x - 3)(x + 2)$
## Step 2: Simplify the expression
$\frac{5(x - 3)}{(x - 3)(x + 2)}=\frac{5}{x + 2}$ (cancel $x - 3$, $x
eq3$)
## Step 3: Find restrictions (denominator ≠ 0)
Original denominator: $(x - 3)(x + 2)
eq0$ ⇒ $x
eq3$ and $x
eq - 2$
# Answer: Simplified: $\frac{5}{x + 2}$, Restrictions: $x
eq3$, $x
eq - 2$

### Question 20
# Explanation:
## Step 1: Factor numerator and denominator
Factor $x^2 - 3x - 10$: $x^2 - 3x - 10=(x - 5)(x + 2)$
Factor $2x - 6$: $2x - 6 = 2(x - 3)$
Factor $8x + 16$: $8x + 16 = 8(x + 2)$
Factor $x^2 - 5x + 6$: $x^2 - 5x + 6=(x - 2)(x - 3)$
## Step 2: Rewrite the expression
$\frac{(x - 5)(x + 2)}{2(x - 3)}\cdot\frac{8(x + 2)}{(x - 2)(x - 3)}$
## Step 3: Simplify
Cancel common factors: $8$ and $2$ give $4$, $(x + 2)$ cancels one, $(x - 3)$ cancels one? Wait, no:
Wait, $\frac{(x - 5)(x + 2)}{2(x - 3)}\cdot\frac{8(x + 2)}{(x - 2)(x - 3)}=\frac{(x - 5)(x + 2)\cdot8(x + 2)}{2(x - 3)\cdot(x - 2)(x - 3)}=\frac{4(x - 5)(x + 2)^2}{(x - 3)^2(x - 2)}$
## Step 4: Find restrictions (denominator ≠ 0)
Original denominators: $2(x - 3)
eq0$ ⇒ $x
eq3$; $(x - 2)(x - 3)
eq0$ ⇒ $x
eq2$ and $x
eq3$; also, numerator factors don't affect, but we check all denominators in original expression. So restrictions: $x
eq2$, $x
eq3$ (from $x - 2 = 0$ and $x - 3 = 0$)
# Answer: Simplified: $\frac{4(x - 5)(x + 2)^2}{(x - 3)^2(x - 2)}$, Restrictions: $x
eq2$, $x
eq3$

### Question 21
# Explanation:
## Step 1: Factor numerator and denominator
Factor $x^2 + 7x - 18$: $x^2 + 7x - 18=(x + 9)(x - 2)$
Factor $x^2 - 81$: $x^2 - 81=(x + 9)(x - 9)$ (difference of squares)
## Step 2: Rewrite the expression
$\frac{x(x - 2)}{(x + 9)(x - 2)}\cdot\frac{(x + 9)(x - 9)}{3(x - 9)(x - 2)}$
## Step 3: Simplify
Cancel common factors: $(x - 2)$ (first numerator and first denominator) cancels (but note $x
eq2$ still), $(x + 9)$ (first denominator and second numerator) cancels, $(x - 9)$ (second numerator and second denominator) cancels. So we have $\frac{x}{1}\cdot\frac{1}{3(x - 2)}=\frac{x}{3(x - 2)}$
## Step 4: Find restrictions (denominator ≠ 0)
Original denominators: $(x + 9)(x - 2)
eq0$ ⇒ $x
eq - 9$, $x
eq2$; $3(x - 9)(x - 2)
eq0$ ⇒ $x
eq9$, $x
eq2$. So restrictions: $x
eq - 9$, $x
eq2$, $x
eq9$
# Answer: Simplified: $\frac{x}{3(x - 2)}$, Restrictions: $x
eq - 9$, $x
eq2$, $x
eq9$

### Question 22
# Explanation:
## Step 1: Rewrite division as multiplication (multiply by reciprocal)
$\frac{x^2 + 10x + 21}{x^2 - 6x - 16}\div\frac{x + 3}{x^2 - 64}=\frac{x^2 + 10x + 21}{x^2 - 6x - 16}\cdot\frac{x^2 - 64}{x + 3}$
## Step 2: Factor numerator and denominator
Factor $x^2 + 10x + 21$: $x^2 + 10x + 21=(x + 3)(x + 7)$
Factor $x^2 - 6x - 16$: $x^2 - 6x - 16=(x - 8)(x + 2)$ (wait, $x^2 - 6x - 16$: looking for two numbers with product -16 and sum -6: -8 and +2, so $(x - 8)(x + 2)$)
Factor $x^2 - 64$: $x^2 - 64=(x + 8)(x - 8)$ (difference of squares)
## Step 3: Simplify
Cancel common factors: $(x + 3)$ (first numerator and second denominator) cancels, $(x - 8)$ (first denominator and second numerator) cancels. So we have $\frac{(x + 7)(x + 8)}{x + 2}$
## Step 4: Find restrictions (denominator ≠ 0)
Original denominators: $x^2 - 6x - 16
eq0$ ⇒ $(x - 8)(x + 2)
eq0$ ⇒ $x
eq8$, $x
eq - 2$; $x + 3
eq0$ ⇒ $x
eq - 3$; $x^2 - 64
eq0$ ⇒ $x
eq\pm8$. So combining, restrictions: $x
eq - 3$, $x
eq - 2$, $x
eq8$ (since $x = 8$ makes $x^2 - 64 = 0$ and $x^2 - 6x - 16 = 0$)
# Answer: Simplified: $\frac{(x + 7)(x + 8)}{x + 2}$, Restrictions: $x
eq - 3$, $x
eq - 2$, $x
eq8$

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QUESTION IMAGE

Question

Explanation:

Question 17

Step 1: Factor numerator and denominator

Step 2: Find restrictions (denominator ≠ 0)

Step 1: Factor numerator and denominator

Step 2: Simplify the expression

Step 3: Find restrictions (denominator ≠ 0)

Step 1: Factor numerator and denominator

Step 2: Simplify the expression

Step 3: Find restrictions (denominator ≠ 0)

Answer:

Question 18