Mastering Multiplying and Dividing Rational Expressions

Rational Expressions
Not in Simplest Form	In Simplest Form
$\frac{x y}{x ( x - 3 ) ( y + 2 )}$	$\frac{x - 1}{x + 1}$
$\frac{x ^{4} + x ^{2}}{x ^{2} + 1}$	$\frac{x ^{3} + 7}{x ^{2} - x - 6}$

Rational Expressions

Not in Simplest Form

In Simplest Form

\frac{x y}{x ( x - 3 ) ( y + 2 )}

\frac{x - 1}{x + 1}

\frac{x ^{4} + x ^{2}}{x ^{2} + 1}

\frac{x ^{3} + 7}{x ^{2} - x - 6}

Expression	Restriction	Excluded Value(s)
$\frac{x - 1}{x + 1}$	$x + 1 \neq = 0$	$x \neq = - 1$
$\frac{x ^{3} + 7}{x ^{2} - x - 6}$	$x^{2} - x - 6 \neq = 0$	$x \neq = - 2$ and $x \neq = 3$
$\frac{x y}{x ( x - 3 ) ( y + 2 )}$	$x (x - 3) (y + 2) \neq = 0$	$x \neq = 0,$ $x \neq = 3,$ and $y \neq = - 2$
$\frac{x ^{4} + x ^{2}}{x ^{2} + 1}$	There is no real number that makes $x^{2} + 1$ zero	None

Expression

Restriction

Excluded Value(s)

\frac{x - 1}{x + 1}

x + 1 \neq = 0

x \neq = - 1

\frac{x ^{3} + 7}{x ^{2} - x - 6}

x^{2} - x - 6 \neq = 0

x \neq = - 2

and

x \neq = 3

\frac{x y}{x ( x - 3 ) ( y + 2 )}

x (x - 3) (y + 2) \neq = 0

x \neq = 0,

x \neq = 3,

and

y \neq = - 2

\frac{x ^{4} + x ^{2}}{x ^{2} + 1}

There is no real number that makes

x^{2} + 1

zero

None

Equivalent Expressions
Rational Expression	Simplified Form
$\frac{x - 3}{( x + 2 ) ( x - 3 )}, x \neq = - 2, 3$	$\frac{1}{x + 2}, x \neq = - 2, 3$
$\frac{x ^{2} + 2 x + 1}{x ^{2} - 1}, x \neq = - 1, 1$	$\frac{x + 1}{x - 1}, x \neq = - 1, 1$
$\frac{x ^{3} - 2 x ^{2} + x}{x ^{2}}, x \neq = 0$	$\frac{x ^{2} - 2 x + 1}{x}, x \neq = 0$

Equivalent Expressions

Rational Expression

Simplified Form

\frac{x - 3}{( x + 2 ) ( x - 3 )}, x \neq = - 2, 3

\frac{1}{x + 2}, x \neq = - 2, 3

\frac{x ^{2} + 2 x + 1}{x ^{2} - 1}, x \neq = - 1, 1

\frac{x + 1}{x - 1}, x \neq = - 1, 1

\frac{x ^{3} - 2 x ^{2} + x}{x ^{2}}, x \neq = 0

\frac{x ^{2} - 2 x + 1}{x}, x \neq = 0

Denominator	Restrictions on the Denominator	Restrictions on the Variable
$(x + 9)^{2}$	$(x + 9)^{2} \neq = 0$	$x \neq = - 9$
$(x + 3) (x - 3)$	$x + 3 \neq = 0$ and $x - 3 \neq = 0$	$x \neq = - 3$ and $x \neq = 3$
$(x + 9) (x + 3)$	$x + 9 \neq = 0$ and $x + 3 \neq = 0$	$x \neq = - 9$ and $x \neq = - 3$

Denominator

Restrictions on the Denominator

Restrictions on the Variable

(x + 9)^{2}

(x + 9)^{2} \neq = 0

x \neq = - 9

(x + 3) (x - 3)

x + 3 \neq = 0

and

x - 3 \neq = 0

x \neq = - 3

and

x \neq = 3

(x + 9) (x + 3)

x + 9 \neq = 0

and

x + 3 \neq = 0

x \neq = - 9

and

x \neq = - 3

	Surface Area, $S$	Volume, $V$
Type I	$\frac{x ^{3} - x ^{2}}{4 x ^{4} + 1 2 x ^{3}}$	$\frac{x ^{2} - 2 x + 1}{x ^{3} + 3 x ^{2}}$
Type II	$\frac{2 x ^{2} - 9 x - 1 8}{4 x ^{2} - 2 8 x + 2 4}$	$2 x^{2} + x - 3$

Surface Area,

S

Volume,

V

Type I

\frac{x ^{3} - x ^{2}}{4 x ^{4} + 1 2 x ^{3}}

\frac{x ^{2} - 2 x + 1}{x ^{3} + 3 x ^{2}}

Type II

\frac{2 x ^{2} - 9 x - 1 8}{4 x ^{2} - 2 8 x + 2 4}

2 x^{2} + x - 3

	Multiplying Rational Expressions
Product	$\frac{2 x ^{2} - 9 x - 1 8}{4 x ^{2} - 2 8 x + 2 4} \cdot \frac{1}{2 x ^{2} + x - 3}$
Factor	$\frac{( 2 x + 3 ) ( x - 6 )}{4 ( x - 1 ) ( x - 6 )} \cdot \frac{1}{( 2 x + 3 ) ( x - 1 )}$
Multiply	$\frac{( 2 x + 3 ) ( x - 6 )}{4 ( x - 1 ) ( x - 6 ) ( 2 x + 3 ) ( x - 1 )}$
Cancel Out Common Factors	$\frac{( 2 x + 3 ) ( x - 6 )}{4 ( x - 1 ) ( x - 6 ) ( 2 x + 3 ) ( x - 1 )}$
Simplify	$\frac{1}{4 ( x - 1 ) ^{2}}$

Multiplying Rational Expressions

Product

\frac{2 x ^{2} - 9 x - 1 8}{4 x ^{2} - 2 8 x + 2 4} \cdot \frac{1}{2 x ^{2} + x - 3}

Factor

\frac{( 2 x + 3 ) ( x - 6 )}{4 ( x - 1 ) ( x - 6 )} \cdot \frac{1}{( 2 x + 3 ) ( x - 1 )}

Multiply

\frac{( 2 x + 3 ) ( x - 6 )}{4 ( x - 1 ) ( x - 6 ) ( 2 x + 3 ) ( x - 1 )}

Cancel Out Common Factors

\frac{( 2 x + 3 ) ( x - 6 )}{4 ( x - 1 ) ( x - 6 ) ( 2 x + 3 ) ( x - 1 )}

Simplify

\frac{1}{4 ( x - 1 ) ^{2}}

	Type I	Type II
Efficiency Ratio	$\frac{x}{4 ( x - 1 )}$	$\frac{1}{4 ( x - 1 ) ^{2}}$
Substitute	$\frac{2}{4 ( 2 - 1 )}$	$\frac{1}{4 ( 2 - 1 ) ^{2}}$
Evaluate	$\frac{1}{2}$	$\frac{1}{4}$

Type I

Type II

Efficiency Ratio

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

\frac{1}{4 ( x - 1 ) ^{2}}

Substitute

\frac{2}{4 ( 2 - 1 )}

\frac{1}{4 ( 2 - 1 ) ^{2}}

Evaluate

\frac{1}{2}

\frac{1}{4}

Find the expression that makes the following statement true.

\frac{x - 3}{x ^{2} + 3 x - 1 8} \div \frac{m m}{x ^{2} + 4 x - 2 1} = \frac{x + 7}{x + 6}

We are asked to find the numerator of the divisor in our equation. We will first first solve an equation for the divisor, and then find its numerator. Let's rearrange the equation to isolate the divisor. Dividend/Divisor= Quotient ⇕ Dividend/Quotient = Divisor Let's apply this fact to rearrange our equation. For simplicity, let Y be the expression we are trying to find. x-3x^2+3x-18/Yx^2+4x-21= x+7/x+6 ⇕ x-3x^2+3x-18/x+7x+6 = Y/x^2+4x-21 Let's now divide the rational expressions on the left-hand side of our new equation. To do so, we will start by factoring the denominator of the dividend.

Let's substitute this expression into our equation. 1/x+7 = Y/x^2+4x-21 In order to solve for Y we can use the Cross Products Property. Let's first factor the denominator of the right-hand side of this equation.

Now we will cross multiply our equation to solve for Y.

We see that the expression x-3 makes the given statement true. x-3/x^2+3x-18 ÷ $x-3$/x^2+4x-21 = x+7/x+6

Perform the indicated operations.

\frac{2 x ^{2} + 5 x - 3}{2 x ^{2} + x - 1 0} \cdot (2 x + 5) \div \frac{2 x - 1}{3 x - 6}

State any restrictions on

x .

We want to multiply and divide the given rational expressions. 2x^2 + 5x - 3/2x^2 + x - 10 * (2x + 5) ÷ 2x - 1/3x - 6 We will begin by factoring the numerators and denominators of each expression if possible, starting with the first.

The second expression cannot be factored further. Let's factor the last expression.

Recall that dividing by a rational expression is the same as multiplying by the reciprocal of the expression. (2x - 1)(x + 3)/(2x + 5)(x - 2) * (2x + 5) ÷ 2x - 1/3(x - 2) ⇕ (2x - 1)(x + 3)/(2x + 5)(x - 2) * (2x + 5) * 3(x - 2)/2x - 1 Let's now multiply all expressions and cancel out any common factors.

We simplified the given expression.

We will identify the restrictions on the variable from the denominator of the original expression and from any other denominator used. For simplicity we will use the factored forms.

Denominator	Restrictions on the Denominator	Restrictions on the Variable
(2x+5)(x-2)	2x+5≠ 0 and x - 2 ≠ 0	x ≠ - 5/2 and x ≠ 2
3x-6	3x-6 ≠ 0	x ≠ 2
(2x+5)(x-2)(2x-1)	2x+5≠ 0, x - 2 ≠ 0, and 2x - 1 ≠ 0	x ≠ - 5/2, x ≠ 2, and x≠ 1/2

We found three unique restrictions on the variable. x≠ - 5/2, x ≠ 1/2, x ≠ 2

Determine whether the given statement is always, sometimes, or never true.

Restrictions on variables change when a rational expression is simplified.

To determine if the given statement is always, sometimes, or never true, let's recall the definition of a simplified rational expression.

Rational Expression in Simple Form |- A rational expression is in simplest form when its numerator and denominator have no common factors.

We factor the numerator and denominator of a rational expression in order to find these common factors. A rational expression and any simplified form must have the same domain in order to be equivalent. Let's see an example.

	Rational Expression	Simplified Form
Expression	x^2-4/x-2	x+2
Domain	x≠ 2	x ≠ 2

As we can see, the restrictions on the variables never change when a rational expression is simplified.

Zosia performs an operation on rational expressions as shown below.

Describe the error in her work.

A . B . C . D . (x + 9) (x - 9) is not the factored form of x^{2} - 81 . The first expression should be multiplied by the reciprocal of the second expression . The factors (x - 3) and (3 - x) do not cancel each other out . The domain should contain - 9 .

We are asked to describe the error in finding the product of two rational expressions. To do so, let's pay close attention to Zosia's work.

In the second line, we see that the factors (x-3) and (3-x) are canceled out. However, the signs of these factors are not the same, so we must factor out a - 1 in one of the factors in order to create common factors in the numerator and denominator. (3-x)= -1 (-3+x) ⇕ (3-x)=-1 (x-3) Let's substitute -1 (x-3) for (3-x) into the second line of Zosia's and cancel out any common factors.

The product of the expressions is equal to - x+9. Also, 3 and - 9 are the excluded values. As a result, the factors (x-3) and (3-x) cannot be canceled out. The answer is C.

Multiplying and Dividing Rational Expressions

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Multiplying and Dividing Rational Expressions

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