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Chapter 7 Rational Functions - College of the Redwoods

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Section 7.2 Reducing <strong>Rational</strong> <strong>Functions</strong> 625<br />

We are led to <strong>the</strong> following key result.<br />

Restrictions. In general, when you reduce a rational expression to lowest terms,<br />

<strong>the</strong> expression obtained should be identical to <strong>the</strong> original expression for all values<br />

<strong>of</strong> <strong>the</strong> variables in each expression, save those values <strong>of</strong> <strong>the</strong> variables that make<br />

any denominator equal to zero. This applies to <strong>the</strong> denominator in <strong>the</strong> original<br />

expression, all intermediate expressions in your work, and <strong>the</strong> final result. We will<br />

refer to any values <strong>of</strong> <strong>the</strong> variable that make any denominator equal to zero as<br />

restrictions.<br />

Let’s look at ano<strong>the</strong>r example.<br />

◮ Example 16.<br />

Reduce <strong>the</strong> expression<br />

to lowest terms. State all restrictions.<br />

2x 2 + 5x − 12<br />

4x 3 + 16x 2 − 9x − 36<br />

(17)<br />

The numerator is a quadratic trinomial with ac = (2)(−12) = −24. The integer<br />

pair −3 and 8 have product −24 and sum 5. Break <strong>the</strong> middle term <strong>of</strong> <strong>the</strong> polynomial<br />

in <strong>the</strong> numerator into a sum using this integer pair, <strong>the</strong>n factor by grouping.<br />

Factor <strong>the</strong> denominator by grouping.<br />

2x 2 + 5x − 12 = 2x 2 − 3x + 8x − 12<br />

= x(2x − 3) + 4(2x − 3)<br />

= (x + 4)(2x − 3)<br />

4x 3 + 16x 2 − 9x − 36 = 4x 2 (x + 4) − 9(x + 4)<br />

= (4x 2 − 9)(x + 4)<br />

= (2x + 3)(2x − 3)(x + 4)<br />

Note how <strong>the</strong> difference <strong>of</strong> two squares pattern was used to factor 4x 2 − 9 = (2x +<br />

3)(2x − 3) in <strong>the</strong> last step.<br />

Now that we’ve factored both numerator and denominator, we cancel common factors.<br />

2x 2 + 5x − 12<br />

4x 3 + 16x 2 − 9x − 36 = (x + 4)(2x − 3)<br />

(2x + 3)(2x − 3)(x + 4)<br />

(x + 4)(2x − 3)<br />

=<br />

(2x + 3)(2x − 3)(x + 4)<br />

= 1<br />

2x + 3<br />

We must now determine <strong>the</strong> restrictions. This means that we must find those values<br />

<strong>of</strong> x that make any denominator equal to zero.<br />

Version: Fall 2007

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