Section1.8Simplifying, multiplying, and dividing rational expressions
ΒΆSubsection1.8.1Written Examples
When presented with a formula for a function named \(f\text{,}\) barring any stated restrictions, the domain of \(f\) is the set of all real numbers for which the formula returns a real number. For example, \(9\) is in the domain of the function \(f(x)=\sqrt{x-5}\) because the value of \(f(9)\) is a real number (specifically \(2\)) where as \(1\) is not in the domain because when we apply the formula to \(1\) the result is \(\sqrt{-4}\) which is not a real number.
A rational function is a function whose formula can be written in the form \(\frac{p(x)}{q(x)}\) where \(p(x)\) and \(q(x)\) are both polynomial functions. Determining the domain of a rational function essentially boils down to determining the values of \(x\) that do not lie in the domain. The values excluded from the domain are the values that result in division by zero (which leads to an undefined result).
To help facilitate identifying the values that cause the denominator to evaluate to zero, we generally completely factor the denominators of rational expressions. For reasons of simplification, which is discussed in the next section, we also completely factor the numerators of rational expression.
Let's determine the domain of the function \(f(x)=\frac{x+3}{x^2-12x+20}\text{.}\) We begin by completely factoring the denominator of the rational expression.
\begin{align*} f(x)\amp=\frac{x+3}{x^2-12x+20}\\ \amp=\frac{x+3}{(x-10)(x-2)} \end{align*}From the factored expression we can see that the value of \(x\) can be neither \(10\) nor \(2\) as either of those values would cause division by \(0\text{.}\) Consequently, the domain of \(f\) consists of all real numbers other that \(2\) and \(10\text{.}\) Using set builder notation we express the domain as \(\{x\mid x \neq 2, x \neq 10\}\) and using interval notation we express the domain as \((-\infty,2) \cup (2,10) \cup (10,\infty)\text{.}\)
Let's determine the domain of the function \(g(t)=\frac{6}{t^2+9t}\text{.}\) We begin by factoring.
\begin{align*} g(t)\amp=\frac{6}{t^2+9t}\\ \amp=\frac{6}{t(t+9)} \end{align*}The input variable for \(g\) is \(t\text{,}\) so we need to determine the values of \(t\) that would result in division by zero. The two values are \(0\) and \(-9\text{,}\) so those are the two values that are not in the domain of \(g\text{.}\) So the domain of \(g\text{,}\) stated using set builder notation, is \(\{x \mid x \neq -9, x \neq 0\}\text{.}\) Using interval notation, the domain of \(g\) is expressed as \((-\infty,-9) \cup (-9,0) \cup (9,\infty)\text{.}\)
One more example. This example requires the factor formula \(a^3-b^3=(a-b)(x^2+ab+b^2)\text{.}\)
Let's determine the domain of \(f(x)=\frac{x-2}{x^3-8}\text{.}\) Factoring we have:
\begin{align*} f(x)\amp=\frac{x-2}{x^3-8}\\ \amp=\frac{x-2}{(x-2)(x^2+2x+4)} \end{align*}If we replace \(x\) with \(2\text{,}\) the resultant expression is \(\frac{0}{0}\text{,}\) which is just as undefined as any other expression involving division by zero. So \(2\) is not in the domain of \(f\text{.}\) It turns out that \(2\) is the only value of \(x\) that results in division by zero. The easiest way to see this is to look at the original denominator. If \(x^3-8=0\text{,}\) then \(x^3=8\text{.}\) The only real number that cubes to \(8\) is \(2\text{,}\) so that is the only number not in the domain of \(f\text{.}\) So the domain of \(f\) is \(\{x \mid x \neq 2\}\) or, equivalently, \((-\infty,2) \cup (2,\infty)\text{.}\)
Recall that multiplying fractions entails multiplying the numerators with one another and multiplying the denominators with one another. For example:
\begin{align*} \frac{2}{7} \cdot \frac{10}{3}\amp=\frac{2 \cdot 10}{7 \cdot 3}\\ \amp=\frac{20}{21} \end{align*}The process of simplifying a rational expression involves employing the multiplication process in reverse, but in a strategic way. What we want to do is completely factor both the numerator and the denominator of the expression and then look for factors that occur in both the numerator and the denominator. We can then factor the common factors from the larger expression as fractions that reduce to one, thereby simplifying the original expression. This will make more sense when seen it in action.
Let's simplify the expression \(\frac{x^2+6x-55}{x^2+x-30}\text{.}\) We begin by factoring both the numerator and the denominator.
\begin{equation*} \frac{x^2+6x-55}{x^2+x-30}=\frac{(x+11)(x-5)}{(x+6)(x-5)} \end{equation*}We now see that there is indeed a factor that is common to both the numerator and the denominator, specifically \((x-5)\text{.}\) We can isolate these factors from the other factors in the form of a fraction that reduces to one. Picking up where we left off:
\begin{align*} \frac{x^2+6x-55}{x^2+x-30}\amp=\frac{(x+11)(x-5)}{(x+6)(x-5)}\\ \amp=\frac{x-5}{x-5} \cdot \frac{x+11}{x+6} \end{align*}So long as the value of \(x\) is not \(5\text{,}\) the expression \(\frac{x-5}{x-5}\) simplifies to \(1\text{.}\) Let's go ahead and replace \(\frac{x-5}{x-5}\) with \(1\text{,}\) but when we do so we have to state a domain restriction, that the value of \(x\) is not allowed to be \(5\text{.}\)
\begin{align*} \frac{x^2+6x-55}{x^2+x-30}\amp=\frac{(x+11)(x-5)}{(x+6)(x-5)}\\ \amp=\frac{x-5}{x-5} \cdot \frac{x+11}{x+6}\\ \amp=1 \cdot \frac{x+11}{x+6}, x \neq 5\\ \amp=\frac{x+11}{x+6}, x \neq 5 \end{align*}As a rule we leave out the line with the explicit factor of \(1\text{.}\) So a complete simplification would be as follows.
\begin{align*} \frac{x^2+6x-55}{x^2+x-30}\amp=\frac{(x+11)(x-5)}{(x+6)(x-5)}\\ \amp=\frac{x-5}{x-5} \cdot \frac{x+11}{x+6}\\ \amp=\frac{x+11}{x+6}, x \neq 5 \end{align*}Several examples follow.
Example
\begin{align*} \frac{t^2+16t+64}{t^2+14t+48}\amp=\frac{(t+8)(t+8)}{(t+8)(t+6)}\\ \amp=\frac{t+8}{t+8} \cdot \frac{t+8}{t+6}\\ \amp=\frac{t+8}{t+6}, t \neq -8 \end{align*}Example
\begin{align*} \frac{x^2-7x}{x^2-15x+56}\amp=\frac{x(x-7)}{(x-8)(x-7)}\\ \amp=\frac{x-7}{x-7} \cdot \frac{x}{x-8}\\ \amp=\frac{x}{x-8}, x \neq 7 \end{align*}Example
\begin{align*} \frac{y^3-36y}{y^3+36y}\amp=\frac{y(y^2-36)}{y(y^2+36)}\\ \amp=\frac{y(y+6)(y-6)}{y(x^2+36)}\\ \amp=\frac{y}{y} \cdot \frac{(y+6)(y-6)}{y^2+36}\\ \amp=\frac{(y+6)(y-6)}{y^2+36}, y \neq 0 \end{align*}As suggested in the last section, multiplying rational expressions entails writing the numerators of the expression over the product of the denominators of the expression. We then simplify the result. Several examples follow.
Example
\begin{align*} \frac{x+3}{x-2} \cdot \frac{x^2-4}{x^2+6x+9}\amp=\frac{(x+3)(x^2-4)}{(x-2)(x^2+6x+9)}\\ \amp=\frac{(x+3)(x+2)(x-2)}{(x-2)(x+3)(x+3)}\\ \amp=\frac{x+3}{x+3} \cdot \frac{x-2}{x-2} \cdot \frac{x+2}{x+3}\\ \amp=\frac{x+2}{x+3}, x \neq 2 \end{align*}Note that we did not need to state the restriction \(x \neq -3\) because a factor of \((x+3)\) is still present in the simplified result.
Example
\begin{align*} \frac{w^3-1}{w^2-1}\amp=\frac{(w-1)(w^2+w+1)}{(w-1)(w+1)}\\ \amp=\frac{w-1}{w-1} \cdot \frac{w^2+w+1}{w+1}\\ \amp=\frac{w^2+w+1}{w+1},w \neq 1 \end{align*}Example
\begin{align*} \frac{10x-35}{5x^2+25x} \cdot \frac{x^2+8x+15}{2x-7}\amp=\frac{(10x-35)(x^2+8x+15)}{(5x^2+25x)(2x-7)}\\ \amp=\frac{5(2x-7)(x+3)(x+5)}{5x(x+5)(2x-7)}\\ \amp=\frac{5}{5} \cdot \frac{2x-7}{2x-7} \cdot \frac{x+5}{x+5} \cdot \frac{x+3}{x}\\ \amp=\frac{x+3}{x}, x \neq \frac{7}{2}, x \neq -5 \end{align*}Example
\begin{align*} \frac{8}{2x+22} \cdot (x^2+22x+121)\amp=\frac{8}{2x+22} \cdot \frac{x^2+22x+121}{1}\\ \amp=\frac{8(x^2+22x+121)}{(2x+22) \cdot 1}\\ \amp=\frac{8(x+11)(x+11)}{2(x+11)}\\ \amp=\frac{x+11}{x+11} \cdot \frac{8}{2} \cdot \frac{x+11}{1}\\ \amp=4(x+11), x \neq -11 \end{align*}Recall that when we divide one fraction by another, we rewrite the quotient as a product after reciprocating the divisor. That is:
\begin{equation*} \frac{a}{b} \div \frac{c}{d}=\frac{a}{b} \cdot \frac{d}{c}\text{.} \end{equation*}When working with rational expressions that include division, we always first rewrite the quotient as a product and then simplify the result. When we convert the quotient to a product, we need to state any domain restrictions in the original divisior that are lost by reciprocating the divisor. For example:
\begin{align*} \frac{3}{x-10} \div \frac{9}{x-5}\amp=\frac{3}{x-10} \cdot \frac{x-5}{9},x \neq 5\\ \amp=\frac{3(x-5)}{(x-10) \cdot 9}, x \neq 5\\ \amp=\frac{3(x-5)}{9(x-10)},x \neq 5 \end{align*}When presented with a quotient in which one of the expressions is a rational expression but the other expression is a polynomial, it can be useful to write the polynomial as a rational expression with a denominator of \(1\text{.}\) Two examples follow.
Example
\begin{align*} \amp(x^2+8x+12) \div \frac{x^2+4x-12}{x^2-36}\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2+8x+12}{1} \div \frac{x^2+4x-12}{x^2-36}\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2+8x+12}{1} \cdot \frac{x^2-36}{x^2+4x-12}, x \neq 6, x \neq -6\\ \amp \phantom{={}} \phantom{={}} =\frac{(x^2+8x+12)(x^2-36)}{1 \cdot (x^2+4x-12)}, x \neq 6, x \neq -6\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+6)(x+2)(x+6)(x-6)}{(x+6)(x-2)},x \neq 6\\ \amp \phantom{={}} \phantom{={}} =\frac{x+6}{x+6} \cdot \frac{(x+2)(x+6)(x-6)}{x-2}, x \neq 6\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+2)(x+6)(x-6)}{x-2}, x \neq 6, x \neq -6 \end{align*}Example
\begin{align*} \frac{x^2-18x+72}{9x^2-16} \div (9x^2-16x)\amp=\frac{x^2-18x+72}{9x^2-16} \div \frac{9x^2-16x}{1}\\ \amp=\frac{x^2-18x+72}{9x^2-16} \cdot \frac{1}{9x^2-16x}\\ \amp=\frac{(x^2-18x+72) \cdot 1}{(9x^2-16)(9x^2-16x)}\\ \amp=\frac{(x-12)(x-6)}{(3x+4)(3x-4) \cdot x(9x-16)}\\ \amp=\frac{(x-12)(x-6)}{x(3x+4)(3x-4)(9x-16)} \end{align*}When there is repeated division between rational expression, every divisor ends up being reciprocated when the expression is rewritten as a product. The is a result of the fact that operations are performed left-to-right. An example follows.
\begin{align*} \amp\frac{x-3}{x-7} \div \frac{x+8}{x-12} \div \frac{x+8}{x-7}\\ \amp \phantom{={}} \phantom{={}} =(\frac{x-3}{x-7} \div \frac{x+8}{x-12}) \div \frac{x+8}{x-7}\\ \amp \phantom{={}} \phantom{={}} =(\frac{x-3}{x-7} \cdot \frac{x-12}{x+8}) \div \frac{x+8}{x-7}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)}{(x-7)(x+8)} \div \frac{x+8}{x-7}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)}{(x-7)(x+8)} \cdot \frac{x-7}{x+8}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)(x-7)}{(x-7)(x+8)(x+8)}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{x-7}{x-7} \cdot \frac{(x-3)(x-12)}{(x+8)(x+8)}, x \neq 12\\ \amp \phantom{={}} \phantom{={}} =\frac{(x-3)(x-12)}{(x+8)^2}, x \neq 12, x \neq 7 \end{align*}Subsection1.8.2Practice Exercises
Determine the domain of each function. State each domain using both set builder notation and interval notation.
- \(f(x)=\frac{2x-8}{x^2-16}\)
- \(g(t)=\frac{t}{t^2+5t-6}\)
- \(h(x)=\frac{4x+8}{2x^2-8x}+\frac{x-3}{x^2-6x+8}\)
- \(m(t)=\frac{t+7}{t^2+49}\)
- \(p(x)=\frac{1}{x^3-8}\)
- \(r(t)=\frac{t-3}{3-t}-\frac{t-2}{t-3}\)
Solution
-
\(\begin{aligned}[t] f(x)\amp=\frac{2x-8}{x^2-16}\\ \amp=\frac{2x-8}{(x-4)(x+4)}\\ \end{aligned}\)
\begin{equation*} \end{equation*}The domain of \(f\) is \(\{x \mid x \neq -4, x \neq 4\}\text{.}\)
The domain of \(f\) is \((-\infty,-4)\cup(-4,4)\cup(4,\infty)\text{.}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] g(t)\amp=\frac{t}{t^2+5t-6}\\ \amp=\frac{t}{(t-6)(t+1)} \end{aligned}\)
\begin{equation*} \end{equation*}The domain of \(g\) is \(\{t \mid t \neq -1, t \neq 6\}\text{.}\)
The domain of \(g\) is \((-\infty,-1) \cup (-1,6) \cup (6,\infty)\text{.}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] h(x)\amp=\frac{4x+8}{2x^2-8x}+\frac{x-3}{x^2-6x+8}\\ \amp=\frac{4x+8}{2x(x-4)}+\frac{x-3}{(x-4)(x-2)} \end{aligned}\)
\begin{equation*} \end{equation*}The domain of \(h\) is \(\{x \mid x \neq 0, x \neq 2, x \neq 4\}\text{.}\)
The domain of \(h\) is \((-\infty,0) \cup (0,2) \cup (2,4) \cup (4,\infty)\text{.}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] m(t)=\frac{t+7}{t^2+49} \end{aligned}\)
\begin{equation*} \end{equation*}The domain of \(m\) is \(\{t \mid t \in \mathbb{R}\}\text{.}\)
The domain of \(m\) is \((-\infty,\infty)\text{.}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] p(x)\amp=\frac{1}{x^3-8}\\ \amp=\frac{1}{(x-2)(x^2+2x+4)} \end{aligned}\)
\begin{equation*} \end{equation*}The domain of \(p\) is \(\{x \mid x \neq 2\}\text{.}\)
The domain of \(p\) is \((-\infty,2) \cup (2,\infty)\text{.}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] r(t)=\frac{t-3}{3-t}-\frac{t-2}{t-3} \end{aligned}\)
\begin{equation*} \end{equation*}The domain of \(r\) is \(\{t \mid t \neq 3\}\text{.}\)
The domain of \(r\) is \((-\infty,3) \cup (3,\infty)\text{.}\)
\begin{equation*} \end{equation*}
Simplify each expression. Make sure that you state any necessary domain restrictions.
- \(\frac{x-3}{x^2+x-12}\)
- \(\frac{x^2-5x+4}{x^2+6x-40}\)
- \(\frac{x^2+7x}{x^2-7x}\)
- \(\frac{x^4+21x^2}{5x^3+x^7}\)
- \(\frac{4t^2-9}{2t-3}\)
- \(\frac{t^2-16}{t^2+16}\)
Solution
-
\(\begin{aligned}[t] \frac{x-3}{x^2+x-12}\amp=\frac{x-3}{(x-3)(x+4)}\\ \amp=\frac{x-3}{x-3} \cdot \frac{1}{x+4}\\ \amp=\frac{1}{x+4}, x \neq 3 \end{aligned}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] \frac{x^2-5x+4}{x^2+6x-40}\amp=\frac{(x-4)(x-1)}{(x-4)(x+10)}\\ \amp=\frac{x-4}{x-4} \cdot \frac{x-1}{x+10}\\ \amp=\frac{x-1}{x+10}, x \neq 4 \end{aligned}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] \frac{x^2+7x}{x^2-7x}\amp=\frac{x(x+7)}{x(x-7)}\\ \amp=\frac{x}{x} \cdot \frac{x+7}{x-7}\\ \amp=\frac{x+7}{x-7}, x \neq 0 \end{aligned}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] \frac{x^4+21x^2}{5x^3+x^7}\amp=\frac{x^2(x^2+21)}{x^3(5+x^4)}\\ \amp=\frac{x^2}{x^2} \cdot \frac{x^2+21}{x(5+x^4)}\\ \amp=\frac{x^2+21}{x(5+x^4)} \end{aligned}\)
\begin{equation*} \end{equation*} -
\(\begin{aligned}[t] \frac{4t^2-9}{2t-3}\amp=\frac{(2t-3)(2t+3)}{2t-3}\\ \amp=\frac{2t-3}{2t-3} \cdot \frac{2t+3}{1}\\ \amp=2t+3, t \neq \frac{3}{2} \end{aligned}\)
\begin{equation*} \end{equation*} \(\begin{aligned}[t] \frac{t^2-16}{t^2+16}=\frac{(t-4)(t+4)}{t^2+16} \end{aligned}\)
Multiply each expression and simplify each result. Make sure that you state any necessary domain restrictions.
- \(\frac{x-4}{2x+3} \cdot \frac{x-1}{x-4}\)
- \(\frac{12x+18}{x^2-1} \cdot \frac{x+1}{6x+9}\)
- \(\frac{4x^2+20x+24}{x^2-25} \cdot \frac{5x+25}{x^2-9}\)
- \((x^2+4x+4) \cdot \frac{x+2}{x^2-4}\)
Solution
\(\begin{aligned}[t] \frac{x-4}{2x+3} \cdot \frac{x-1}{x-4}\amp=\frac{(x-4)(x-1)}{(2x+3)(x-4)}\\ \amp=\frac{x-4}{x-4} \cdot \frac{x-1}{2x+3}\\ \amp=\frac{x-1}{2x+3}, x \neq 4 \end{aligned}\)
\(\begin{aligned}[t] \frac{12x+18}{x^2-1} \cdot \frac{x+1}{6x+9}\amp=\frac{(12x+18)(x+1)}{(x^2-1)(6x+9)}\\ \amp=\frac{6(2x+3)(x+1)}{(x-1)(x+1) \cdot 3(2x+3)}\\ \amp=\frac{x+1}{x+1} \cdot \frac{2x+3}{2x+3} \cdot \frac{6}{3(x-1)}\\ \amp=\frac{2}{x-1}, x \neq -1, x \neq -\frac{3}{2} \end{aligned}\)
\(\begin{aligned}[t] \frac{4x^2+20x+24}{x^2-25} \cdot \frac{5x+25}{x^2-9}\amp=\frac{(4x^2+20x+24)(5x+25)}{(x^2-25)(x^2-9)}\\ \amp=\frac{4(x^2-5x+6) \cdot 5(x+5)}{(x-5)(x+5)(x-3)(x+3)}\\ \amp=\frac{20(x-3)(x-2)(x+5)}{(x-5)(x+5)(x-3)(x+3)}\\ \amp=\frac{x-3}{x-3} \cdot \frac{x+5}{x+5} \cdot \frac{20(x-3)}{(x-5)(x+3)}\\ \amp=\frac{20(x-3)}{(x-5)(x+3)}, x \neq 3, x \neq -5 \end{aligned}\)
-
\(\begin{aligned}[t] (x^2+4x+4) \cdot \frac{x+2}{x^2-4}\amp=\frac{x^2+4x+4}{1} \cdot \frac{x+2}{x^2-4}\\ \amp=\frac{(x^2+4x+4)(x+2)}{x^2-4}\\ \amp=\frac{(x+2)(x+2)(x+2)}{(x+2)(x-2)}\\ \amp=\frac{x+2}{x+2} \cdot \frac{(x+2)(x+2)}{x-2}\\ \amp=\frac{(x+2)^2}{x-2}, x \neq -2 \end{aligned}\)
\begin{equation*} \end{equation*}
Multiply and/or divide each expression and simplify each result. Make sure that you state any necessary domain restrictions.
- \(\frac{x^2-9}{x+1} \div \frac{x+3}{x^2-1}\)
- \(\frac{x+5}{x^2-6x-7} \cdot \frac{x^2+1}{x^2+11x+30} \div \frac{x-7}{x+1}\)
- \(\frac{x^2+2x-15}{4x+20} \div (x-3)\)
- \(\frac{2x}{x-2} \div \frac {x+2}{x} \div \frac{7x}{x^2-4}\)
Solution
- \(\begin{aligned}[t] \amp\frac{x^2-9}{x+1} \div \frac{x+3}{x^2-1}\\ \amp \phantom{={}} \phantom{={}} =\frac{x^2-9}{x+1} \cdot \frac{x^2-1}{x+3}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x^2-9)(x^2-1)}{(x+1)(x+3)}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+3)(x-3)(x+1)(x-1)}{(x+1)(x+3)}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =\frac{x+3}{x+3} \cdot \frac{x+1}{x+1} \cdot \frac{(x-3)(x-1)}{1}, x \neq 1\\ \amp \phantom{={}} \phantom{={}} =(x-3)(x-1), x \neq 1, x \neq -1, x \neq -3 \end{aligned}\) \begin{equation*} \end{equation*}
- \(\begin{aligned}[t] \amp\frac{x+5}{x^2-6x-7} \cdot \frac{x^2+1}{x^2+11x+30} \div \frac{x-7}{x+1}\\ \amp \phantom{={}} \phantom{={}} =\frac{x+5}{x^2-6x-7} \cdot \frac{x^2+1}{x^2+11x+30} \cdot \frac{x+1}{x-7}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+5)(x^2+1)(x+1)}{(x^2-6x+7)(x^2+11x+30)(x-7)}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x+5)(x^2+1)(x+1)}{(x+7)(x-1)(x+5)(x+6)(x-7)}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{x+5}{x+5} \cdot \frac{(x^2+1)(x+1)}{(x-1)(x+6)(x-7)}, x \neq -1\\ \amp \phantom{={}} \phantom{={}} =\frac{(x^2+1)(x+1)}{(x-1)(x+6)(x-7)}, x \neq -1, x \neq -5 \end{aligned}\) \begin{equation*} \end{equation*}
- \(\begin{aligned}[t] \frac{x^2+2x-15}{4x+20} \div (x-3)\amp=\frac{x^2+2x-15}{4x+20} \div \frac{x-3}{1}\\ \amp=\frac{x^2+2x-15}{4x+20} \cdot \frac{1}{x-3}\\ \amp=\frac{x^2+2x-15}{(4x+20)(x-3)}\\ \amp=\frac{(x+5)(x-3)}{4(x+5)(x-3)}\\ \amp=\frac{x+5}{x+5} \cdot \frac{x-3}{x-3} \cdot \frac{1}{4}\\ \amp=\frac{1}{4}, x \neq -5, x \neq 3 \end{aligned}\) \begin{equation*} \end{equation*}
- \(\begin{aligned}[t] \amp\frac{2x}{x-2} \div \frac{x+2}{x} \div \frac{7x}{x^2-4}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x}{x-2} \cdot \frac {x}{x+2} \cdot \frac{x^2-4}{7x}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x \cdot x \cdot (x^2-4)}{(x-2) \cdot (x+2) \cdot 7x}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x^2(x+2)(x-2)}{7x(x-2)(x+2)}\\ \amp \phantom{={}} \phantom{={}} =\frac{x}{x} \cdot \frac{x+2}{x+2} \cdot \frac{x-2}{x-2} \cdot \frac{2x}{7}\\ \amp \phantom{={}} \phantom{={}} =\frac{2x}{7}, x \neq 0, x \neq 2, x \neq -2 \end{aligned}\)