Explanation Given: The rational expression is, 3 k 2 k 2 + 3 k − 2 − 2 k 2 k 2 − 7 k + 3 . Property used: Fundamental property, P Q = P S Q S , for all mathematical expressions P , Q , R , and S , with Q ≠ 0 and S ≠ 0 . Subtraction property, P Q − R Q = P − R Q , for all mathematical expressions P , Q , and R , with Q ≠ 0 . Calculation: Since the denominators of the two rational expressions are different, take least common denominator and then add the numerator while keeping the same denominator. Obtain the factors of the denominators. 2 k 2 + 3 k − 2 = 2 k 2 − k + 4 k − 2 = k ( 2 k − 1 ) + 2 ( 2 k − 1 ) = ( 2 k − 1 ) ( k + 2 ) 2 k 2 − 7 k + 3 = 2 k 2 − k − 6 k + 3 = k ( 2 k − 1 ) − 3 ( 2 k − 1 ) = ( 2 k − 1 ) ( k − 3 ) Thus, the rational expression becomes, 3 k ( 2 k − 1 ) ( k + 2 ) − 2 k ( 2 k − 1 ) ( k − 3 ) . The least common denominator of ( 2 k − 1 ) ( k + 2 ) and ( 2 k − 1 ) ( k − 3 ) is ( k + 2 ) ( 2 k − 1 ) ( k − 3 ) . Multiply the first term by ( x − 2 ) ( x − 2 ) and the second term by ( x + 3 ) ( x + 3 ) and apply Fundamental property as follows. 3 k 2 k 2 + 3 k − 2 − 2 k 2 k 2 − 7 k + 3 = 3 k ( 2 k − 1 ) ( k + 2 ) − 2 k ( 2 k − 1 ) ( k − 3 ) = 3 k ( 2 k − 1 ) ( k + 2 ) ⋅ ( k − 3 ) ( k − 3 ) − 2 k ( 2 k − 1 ) ( k − 3 ) ⋅ ( k + 2 ) ( k + 2 )

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter R.3, Problem 35E

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To express: The rational expression 3k2k2+3k−2−2k2k2−7k+3 in lowest terms.

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Chapter R.3, Problem 34E

Chapter R.3, Problem 36E

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To express: The rational expression 3 k 2 k 2 + 3 k − 2 − 2 k 2 k 2 − 7 k + 3 in lowest terms.

Solution Summary: The author explains the rational expression, 2k2, in lowest terms, with the least common denominator and subtraction property.

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To express: The rational expression 3k2k2+3k−2−2k2k2−7k+3 in lowest terms.

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Chapter R Solutions