rational_expressions

rational_expressions - Rational expressions Preliminaries....

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Rational expressions Preliminaries . Recall that a division of real numbers a b , where b 0, can be performed by multiplying the real number a by the multiplicative inverse = a ( ) - 1 1 a of a . Thus = a b a . 1 b . Consider a rational expression of the form a 1 . a 2 . . . . . a m b 1 . b 2 . . . . . b n , where a 1 , a 2 , . . . a m , b 1 , b 2 , . . . b n are real numbers. The notation suggests that the evaluation of this expression is to be performed in the order suggested by the expression: ( ) a 1 . a 2 . . . . . a m / ( ) b 1 . b 2 . . . . . b n , that is, the values for the products in the numerator and denominator are computed independently and then a single division of the numerator by the denominator produces the final value. Thus 2 . 3 . 4 . 5 6 . 7 . 8 . 9 . 2 = 120 3024 2 = 120 3024 . 1 2 = 5 126 . 1 2 = 5 2 126 . However, a 1 . a 2 . . . . . a m b 1 . b 2 . . . . . b n = a 1 . a 2 . . . . . a m . 1 b 1 . b 2 . . . . . b n . = a 1 . a 2 . . . . . a m . 1 b 1 . 1 b 2 . . . . . 1 b n . Making use of the commutative and associative properties of multiplication this gives many ways of computing the value of such an expression. For example, 2 . 3 . 4 . 5 6 . 7 . 8 . 9 . 2 = 2 . 3 6 . 4 8 . 5 7 . 9 . 1 2 = 1 . 1 2 . 5 63 . 1 2 = 5 126 2 . This result can be acheived by the standard method of "cancelling". The previous calculation may be written as follows. 2 . 3 . 4 . 5 6 . 7 . 8 . 9 . 2 = 4 . 5 7 . 8 . 9 . 2 = 5 7 . 2 . 9 . 2 = 5 126 2 .
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Rational expressions with polynomials in the numerator and denominator . Rational expressions with polynomials in the numerator and denominator can sometimes be simplified by factoring the polynomials. Example 1 : (a) Simplify the expression - - + 2 x 3 3 x 2 8 x 12 - + 4 x 3 14 x 2 12 x . (b) If possible, find the value of the expression in part (a) when = x 5 and also when = x 3 2 . Solution : (a) - - + 2 x 3 3 x 2 8 x 12 - + 4 x 3 14 x 2 12 x = - x 2 ( ) - 2 x 3 4 ( ) - 2 x 3 2 x ( ) - + 2 x 2 7 x 6 = ( ) - x 2 4 ( ) - 2 x 3 2 x ( ) - 2 x 3 ( ) - x 2 = ( ) - x 2 ( ) + x 2 ( ) - 2 x 3 2 x ( ) - 2 x 3 ( ) - x 2 = + x 2 2 x . ____ The last step can be broken up as follows. = ( ) - x 2 ( ) + x 2 ( ) - 2 x 3 2 x ( ) - 2 x 3 ( ) - x 2 + x 2 2 x . - x 2 - x 2 . - 2 x 3 - 2 x 3 = + x 2 2 x . 1 . 1 = + x 2 2 x . Replacing - x 2 - x 2 by 1 is usually regarded as "cancelling" the ( ) - x 2 factor in the numerator of the expression with the factor ( ) - x 2 in the denominator. However this is not correct to replace - x 2 - x 2 by 1 when = x 2 because, in this case, - x 2 - x 2 becomes 0 0 , which is a meaningless expression . If a b is to be understood to mean a . 1 b , then = 0 0 0 . 1 0 , but 1 0 does not exist as a real number. Similarly, it is not correct to replace - 2 x 3 - 2 x 3 by 1 when = x 3 2 , because - 2 x 3 - 2 x 3 becomes 0 0 when = x 3 2 . Hence, the previous simplification is not valid when
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This note was uploaded on 11/12/2011 for the course MATH 111 taught by Professor Uri during the Spring '08 term at Rutgers.

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rational_expressions - Rational expressions Preliminaries....

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