McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Multiplying and Dividing Rational Expressions
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Exercise 35 Page 534

Simplify the expressions in the numerator and those in the denominator. Then, multiply the new numerator by the reciprocal of the new denominator.

Quotient: (x-3)(x+1)/6(x+7)
Restrictions: x≠ - 7, x≠ -3, x ≠ -1, x≠ 2

Practice makes perfect

A complex fraction is a rational expression that has at least one fraction in its numerator, denominator, or both. We want to simplify the given complex fraction. x^2-96x-12/x^2+10x+21x^2-x-2 To do so, we will simplify the expressions in the numerator and those in the denominator. Then, we will multiply the new numerator by the reciprocal of the new denominator. Let's start with simplifying the numerator.

x^2-9/6x-12
x^2-3^2/6x-12
(x+3)(x-3)/6x-12
(x+3)(x-3)/6(x-2)
Next, we will simplify the denominator of the complex fraction.

x^2+10x+21/x^2-x-2
â–¼
Factor the numerator
x^2+3x+7x+21/x^2-x-2
x(x+3)+7(x+3)/x^2-x-2
(x+3)(x+7)/x^2-x-2
â–¼
Factor the denominator
(x+3)(x+7)/x^2+x-2x-2
(x+3)(x+7)/x(x+1)-2(x+1)
(x+3)(x+7)/(x+1)(x-2)

Now, we can rewrite the complex fraction. x^2-96x-12/x^2+10x+21x^2-x-2 ⇔ (x+3)(x-3)6(x-2)/(x+3)(x+7)(x+1)(x-2) Finally, we will multiply the numerator by the reciprocal of the new denominator. (x+3)(x-3)6(x-2)/(x+3)(x+7)(x+1)(x-2) ⇔ (x+3)(x-3)/6(x-2) * (x+1)(x-2)/(x+3)(x+7) Let's simplify the above expression.

(x+3)(x-3)/6(x-2) * (x+1)(x-2)/(x+3)(x+7)
(x+3)(x-3)/6(x-2) * (x+1)(x-2)/(x+3)(x+7)
(x-3)(x+1)/6(x+7)

We will now identify the restrictions from the denominator of the simplified expression and from any other denominator used. For simplicity, we will use their factored forms.

Denominator Restrictions on the denominator Restrictions on the variable
6(x-2) x-2≠ 0 x≠ 2
(x+1)(x-2) x+1≠ 0 and x-2≠ 0 x≠ - 1 and x ≠ 2
(x+3)(x+7) x+3≠ 0 and x+7≠ 0 x≠ -3 and x ≠ -7
6(x+7) x+7≠ 0 x ≠ -7

We found four restrictions on the variable. x≠ - 7, x≠ -3, x ≠ -1, and x≠ 2