Question

find the quotients and any excluded values.

1. $\frac{5x^{2}+10x}{x^{2}+2x + 1}div\frac{20x + 40}{x^{2}-1}$
2. $\frac{x^{2}-9x + 18}{x^{2}+9x + 18}div\frac{x^{2}-36}{x^{2}-9}$
3. $\frac{-x^{2}+x + 20}{5x^{2}-25x}div\frac{x + 4}{2x - 14}$
4. $\frac{x + 3}{x^{2}+8x + 15}div\frac{x^{2}-25}{x - 5}$
5. $\frac{x^{2}-10x + 9}{3x}div\frac{x^{2}-7x - 18}{x^{2}+2x}$
6. $\frac{8x + 32}{x^{2}+8x + 16}div\frac{x^{2}-6x}{x^{2}-2x - 24}$

let $p(x)=\frac{1}{x + 1}$ and $q(x)=\frac{1}{x - 1}$. perform the operation, and show that it results in another rational expression.

1. $p(x)+q(x)$
2. $p(x)-q(x)$

Explanation:

Step1: Rewrite division as multiplication

$\frac{5x^2 + 10x}{x^2 + 2x + 1} \div \frac{20x + 40}{x^2 - 1} = \frac{5x^2 + 10x}{x^2 + 2x + 1} \times \frac{x^2 - 1}{20x + 40}$

Step2: Factor all polynomials

$\frac{5x(x+2)}{(x+1)^2} \times \frac{(x-1)(x+1)}{20(x+2)}$

Step3: Cancel common factors

$\frac{5x\cancel{(x+2)}}{(x+1)\cancel{(x+1)}} \times \frac{(x-1)\cancel{(x+1)}}{20\cancel{(x+2)}} = \frac{5x(x-1)}{20(x+1)}$

Step4: Simplify the constant

$\frac{x(x-1)}{4(x+1)}$

Step5: Find excluded values

Denominators cannot be 0: $x+1
eq 0 \implies x
eq -1$; $x+2
eq 0 \implies x
eq -2$; $x-1
eq 0 \implies x
eq 1$

---

Step1: Rewrite division as multiplication

$\frac{x^2 - 9x + 18}{x^2 + 9x + 18} \div \frac{x^2 - 36}{x^2 - 9} = \frac{x^2 - 9x + 18}{x^2 + 9x + 18} \times \frac{x^2 - 9}{x^2 - 36}$

Step2: Factor all polynomials

$\frac{(x-3)(x-6)}{(x+3)(x+6)} \times \frac{(x-3)(x+3)}{(x-6)(x+6)}$

Step3: Cancel common factors

$\frac{(x-3)\cancel{(x-6)}}{\cancel{(x+3)}(x+6)} \times \frac{(x-3)\cancel{(x+3)}}{\cancel{(x-6)}(x+6)} = \frac{(x-3)^2}{(x+6)^2}$

Step4: Find excluded values

Denominators cannot be 0: $x+3
eq 0 \implies x
eq -3$; $x+6
eq 0 \implies x
eq -6$; $x-6
eq 0 \implies x
eq 6$; $x-3
eq 0 \implies x
eq 3$

---

Step1: Rewrite division as multiplication

$\frac{-x^2 + x + 20}{5x^2 - 25x} \div \frac{x+4}{2x-14} = \frac{-x^2 + x + 20}{5x^2 - 25x} \times \frac{2x-14}{x+4}$

Step2: Factor all polynomials

$\frac{-(x^2 - x - 20)}{5x(x-5)} \times \frac{2(x-7)}{x+4} = \frac{-(x-5)(x+4)}{5x(x-5)} \times \frac{2(x-7)}{x+4}$

Step3: Cancel common factors

$\frac{-\cancel{(x-5)}\cancel{(x+4)}}{5x\cancel{(x-5)}} \times \frac{2(x-7)}{\cancel{(x+4)}} = -\frac{2(x-7)}{5x}$

Step4: Find excluded values

Denominators cannot be 0: $x
eq 0$; $x-5
eq 0 \implies x
eq 5$; $x+4
eq 0 \implies x
eq -4$; $x-7
eq 0 \implies x
eq 7$

---

Step1: Rewrite division as multiplication

$\frac{x+3}{x^2 + 8x + 15} \div \frac{x^2 -25}{x-5} = \frac{x+3}{x^2 + 8x + 15} \times \frac{x-5}{x^2 -25}$

Step2: Factor all polynomials

$\frac{x+3}{(x+3)(x+5)} \times \frac{x-5}{(x-5)(x+5)}$

Step3: Cancel common factors

$\frac{\cancel{x+3}}{\cancel{(x+3)}(x+5)} \times \frac{\cancel{x-5}}{\cancel{(x-5)}(x+5)} = \frac{1}{(x+5)^2}$

Step4: Find excluded values

Denominators cannot be 0: $x+3
eq 0 \implies x
eq -3$; $x+5
eq 0 \implies x
eq -5$; $x-5
eq 0 \implies x
eq 5$

---

Step1: Rewrite division as multiplication

$\frac{x^2 -10x +9}{3x} \div \frac{x^2 -7x -18}{x^2 +2x} = \frac{x^2 -10x +9}{3x} \times \frac{x^2 +2x}{x^2 -7x -18}$

Step2: Factor all polynomials

$\frac{(x-1)(x-9)}{3x} \times \frac{x(x+2)}{(x-9)(x+2)}$

Step3: Cancel common factors

$\frac{(x-1)\cancel{(x-9)}}{3\cancel{x}} \times \frac{\cancel{x}\cancel{(x+2)}}{\cancel{(x-9)}\cancel{(x+2)}} = \frac{x-1}{3}$

Step4: Find excluded values

Denominators cannot be 0: $x
eq 0$; $x-9
eq 0 \implies x
eq 9$; $x+2
eq 0 \implies x
eq -2$; $x-1
eq 0 \implies x
eq 1$

---

Step1: Rewrite division as multiplication

$\frac{8x+32}{x^2 +8x +16} \div \frac{x^2 -6x}{x^2 -2x -24} = \frac{8x+32}{x^2 +8x +16} \times \frac{x^2 -2x -24}{x^2 -6x}$

Step2: Factor all polynomials

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

Step3: Cancel common factors

$\frac{8\cancel{(x+4)}}{\cancel{(x+4)}\cancel{(x+4)}} \times \frac{\cancel{(x-6)}\cancel{(x+4)}}{x\cancel{(x-6)}} = \frac{8}{x}$

Step4: Find excluded values

Denominators cannot be 0: $x+4
eq 0 \implies x
eq -4$; $x
eq 0$; $x-6
eq 0 \implies x
eq 6$

---

Step1: Find common denominator

$p(x)+q(x) = \frac{1}{x+1} + \frac{1}{x-1} = \frac{(x-1)+(x+1)}{(x+1)(x-1)}$

Step2: Simplify numerator

$\frac{x-1+x+1}{x^2 -1} = \frac{2x}{x^2 -1}$

Step3: Verify rational form

This is a ratio of polynomials, so it is rational. Excluded values: $x
eq \pm1$

---

Step1: Find common denominator

$p(x)-q(x) = \frac{1}{x+1} - \frac{1}{x-1} = \frac{(x-1)-(x+1)}{(x+1)(x-1)}$

Step2: Simplify numerator

$\frac{x-1-x-1}{x^2 -1} = \frac{-2}{x^2 -1}$

Step3: Verify rational form

This is a ratio of polynomials, so it is rational. Excluded values: $x
eq \pm1$

Answer:

1. Quotient: $\frac{x(x-1)}{4(x+1)}$; Excluded values: $x

eq -2, -1, 1$

1. Quotient: $\frac{(x-3)^2}{(x+6)^2}$; Excluded values: $x

eq -6, -3, 3, 6$

1. Quotient: $-\frac{2(x-7)}{5x}$; Excluded values: $x

eq -4, 0, 5, 7$

1. Quotient: $\frac{1}{(x+5)^2}$; Excluded values: $x

eq -5, -3, 5$

1. Quotient: $\frac{x-1}{3}$; Excluded values: $x

eq -2, 0, 1, 9$

1. Quotient: $\frac{8}{x}$; Excluded values: $x

eq -4, 0, 6$

1. $p(x)+q(x)=\frac{2x}{x^2 -1}$, which is a rational expression (ratio of polynomials); Excluded values: $x

eq \pm1$

1. $p(x)-q(x)=\frac{-2}{x^2 -1}$, which is a rational expression (ratio of polynomials); Excluded values: $x

eq \pm1$