73 R.7 Rational Expressions EXAMPLE 2 Writing Rational Expressions in LowestTerms Write each rational expression in lowest terms. (a) 2x2 + 7x - 4 5x2 + 20x (b) 6 - 3x x2 - 4 SOLUTION (a) 2x2 + 7x - 4 5x2 + 20x = 12x - 121x + 42 5x1x + 42 Factor. = 2x - 1 5x Divide out the common factor. To determine the domain, we find values of x that make the original denominator 5x2 + 20x equal to 0, and exclude them. 5x2 + 20x = 0 Set the denominator equal to 0. 5x1x + 42 = 0 Factor. 5x = 0 or x + 4 = 0 Zero-factor property x = 0 or x = -4 Solve each equation. The domain is 5x x ≠0, -46. From now on, we will assume such restrictions when writing rational expressions in lowest terms. (b) 6 - 3x x2 - 4 = 312 - x2 1x + 221x - 22 Factor. = 312 - x21-12 1x + 221x - 221-12 2 - x and x - 2 are opposites. Multiply numerator and denominator by -1. = 312 - x21-12 1x + 2212 - x2 1x - 221-12 = -x + 2 = 2 - x = -3 x + 2 Divide out the common factor. Working in an alternative way would lead to the equivalent result 3 -x - 2 . S Now Try Exercises 23 and 27. Be careful with signs. CAUTION The fundamental principle requires a pair of common factors, one in the numerator and one in the denominator. Only after a rational expression has been factored can any common factors be divided out. For example, 2x + 4 6 = 21x + 22 2 # 3 = x + 2 3 . Factor first, and then divide. LOOKING AHEAD TO CALCULUS A standard problem in calculus is investigating what value an expression such as x 2 - 1 x - 1 approaches as x approaches 1. We cannot do this by simply substituting 1 for x in the expression since the result is the indeterminate form 0 0. When we factor the numerator and write the expression in lowest terms, it becomes x + 1. Then, by substituting 1 for x, we obtain 1 + 1 = 2, which is the limit of x 2 - 1 x - 1 as x approaches 1.
RkJQdWJsaXNoZXIy NjM5ODQ=