Explanation Given: The expression is 3 a a 2 + 5 a − 6 − 2 a a 2 + 7 a + 6 . Procedure used: Adding and Subtracting Fractions For fractions a b and c d ( b ≠ 0 , d ≠ 0 ) , the sum of the fractions is a b + c d = a d + b c b d and the difference of the fractions is a b − c d = a d − b c b d . Finding the Least Common Denominator ( LCD ) Step1 Write each denominator as a product of prime factors. Step2 Form a product of all the different prime factors. Each factor should have an exponent the greatest exponent that appears on that factor.” Calculation: The given expression is 3 a a 2 + 5 a − 6 − 2 a a 2 + 7 a + 6 . Find the LCD of the denominators ( a 2 + 5 a − 6 ) and ( a 2 + 7 a + 6 ) . Step1: Write each denominator as a product of prime factors. ( a 2 + 5 a − 6 ) = ( a + 6 ) ( a − 1 ) ( a 2 + 7 a + 6 ) = ( a + 6 ) ( a − 1 ) Step2: Obtain the product of the factors such that each factor is multiplied the greatest number of times it occurs in either number. LCD = ( a + 6 ) 1 ⋅ ( a − 1 ) 1 ⋅ ( a + 1 ) 1 = ( a + 6 ) ( a − 1 ) ( a + 1 ) . Now, simplify the expression 3 a a 2 + 5 a − 6 − 2 a a 2 + 7 a + 6 as follows. 3 a a 2 + 5 a − 6 − 2 a a 2 + 7 a + 6 = ( 3 a ) ⋅ ( a + 1 ) ( a + 6 ) ( a − 1 ) ⋅ ( a + 1 ) − ( 2 a ) ⋅ ( a − 1 ) ( a + 6 ) ( a + 1 ) ⋅ ( a − 1 ) ( LCD = ( a +

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ISBN: 8220100802669

Author: Rockswold

Publisher: PEARSON

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Chapter R.3, Problem 61E

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The difference of the expressions 3aa2+5a−6 and 2aa2+7a+6.

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Chapter R.3, Problem 60E

Chapter R.3, Problem 62E

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Chapter R Solutions

EBK GRAPHICAL APPROACH TO COLLEGE ALGEB

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The difference of the expressions 3 a a 2 + 5 a − 6 and 2 a a 2 + 7 a + 6 .

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The difference of the expressions 3aa2+5a−6 and 2aa2+7a+6.

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Chapter R Solutions