Addition and Subtraction

Transcription

1 Addition nd Subtrction Nme: Dte: Definition: rtionl expression A rtionl expression is n lgebric expression in frction form, with polynomils in the numertor nd denomintor such tht t lest one vrible ppers in the denomintor. e.g. x or 2x 2 + x 3 +4x 2 3x Definition: domin of n lgebric expression The domin of n lgebric expression is the set of rel numbers tht the vrible is permitted to hve. It my be helpful to think of the domin s the diet of the expression. Exercise D: Determine the domin of ech of the following expressions. ) x 4 2 b) 4p x c) x x +3 ) Solution: Notice tht x 4 2 x x 2. This is liner polynomil! There re no domin/dietry restrictions for polynomils. Therefore, the domin is given by the complete set of rel numbers, R. b) Solution: Recll tht whenever we hve n nth root where the n is even, the expression under the rdicl must be nonnegtive. For the expression 4p x, n 4. Thus, the domin restrictions re: x 6< 0. In other words, the domin is: x 2 [0, ) c) Solution: Here, we hve rtionl expression. Becuse the numertor is polynomil, there re no restrictions here. For the denomintor, we require tht x Therefore, the domin is the set of rel numbers such tht x In other words, x 2 R such tht x 6 3 or equivlently, x 2 (, 3) [ ( 3, ). Exercise D2: Determine if the following sttement is TRUE or FALSE. Solution: FALSE. We know tht if is some rel number, where 6 0, then x+2. Recll tht the rtionl expression x+2 chnges s x chnges. Furthermore, in the cse where x 2, we hve This is n indeterminte form, which does not lwys equl. The correct sttement is: x +2, provided x 6 2 x +2 x+2 x+2 Mking Mth Possible of 6 c S diyy Hendrickson

2 Addition nd Subtrction. Suppose you were sked to simplify the following rtionl expression: x + 6 x 2 +2x + + x 2 Gol # Understnding the importnce of n LCD (Lest Common Denomintor) When ttempting to simplify n lgebric expression, it my be helpful to recll our pproch with numericl exmples. We know tht in order to dd or subtrct frctions, we must first hve common denomintor. Below is common property to help us chieve this gol: b ± c d d ± bc bd (Property ) Although this property gurntees tht we get common denomintor, we my or my not hve found the Lest Common Denomintor (LCD). So, in moment, we will explore how to find it since LCDs hve the potentil to drsticlly simplify our computtions lter on. Consider the following exmple: 2 8 Using Property, we hve: 2 8 (8) (2) 2(8) Notice tht 26 is six times lrger thn the reduced denomintor of 36! If you found it obvious tht the LCD ws 36, you re o to gret strt! Gol # 2 Finding LCDs Systemticlly Note tht LCDs re not lwys esy to identify. So, we will explore more systemtic pproch to finding them! If b nd d denote denomintors, we hve the following cses: Cse I: One denomintor is fctor of the other. e.g. 2 3 ) LCD Lrger denomintor In the exmple, 3 is fctor of 8, so the LCD8! Cse II: The gretest common fctor of the denomintors is. i.e. GCD (b, d) This mens tht the denomintors hve no common prime fctors. e.g ) LCD b d (the product of the two denomintors) In the exmple, nd Therefore, the LCD ! 8 Mking Mth Possible 2 of 6 c S diyy Hendrickson

3 Addition nd Subtrction Cse III: The denomintors hve prime fctors in common. e.g. 2 In this cse, we hve set of strtegies for finding the LCD! 8 S Fctor ech denomintor (D nd D 2 ). In other words, express D nd D 2 s product of their prime fctors (i.e. the prts tht mke them). Why? So tht we cn esily see the minimum requirements needed in order to include every denomintor in the LCD. For exmple: D : D 2 : S2 Construct the LCD by multiplying together the highest power of ech prime tht ppers mong the denomintors. Why? We know tht every denomintor must be fctor of the LCD, so by tking the highest exponent on ech prime, we gurntee tht no denomintor will end up missing some of its prts. For exmple, if we include 2 2,wewillcoverthe requirement for 2 for both D nd D 2. However, if we hd only tke 2, only D 2 will be stisfied by the LCD. LCD S3 Multiply ech frction s numertor nd denomintor by the missing fctors needed to crete the LCD, then simplify the expression. Why? Remember tht we cnnot chnge the originl problem. So, wht we do to the bottom, we must lwys do to the top, which simply mens we re multiplying by. Our gol ws just to mnipulte the expression so tht we hve n LCD (3) (2) Remember: Every strtegy hs purpose! Mking Mth Possible 3 of 6 c S diyy Hendrickson

4 Addition nd Subtrction Gol #3 Becuse vribles re just numbers in disguise, we cn use the sme strtegic pproch (bove) for the rtionl expression first proposed: x + 6 x 2 +2x + + x 2 S Fctor the denomintors. D : (x + ) (A prime polynomil fctor) D 2 : x 2 +2x +(x + ) 2 (Perfect Squre; requires fctoring) D 3 : x 2 (x + )(x ) (Di erence of Squres; requires fctoring) S2 Construct the LCD by multiplying together the highest power of ech fctor tht ppers mong the denomintors LCD (x + ) 2 (x ) Q: Wht would our denomintor hve been if we hd used Propery? A: (x + )(x 2 +2x + )(x 2 )... polynomil of degree insted of 3, which lso mens lot more work when simplifying the numertor! S3 Multiply ech frction s numertor nd denomintor by the missing fctors needed to crete the LCD, then simplify the expression. Be sure to stte restrictions before cncelling. This is importnt now tht our expressions involve vribles! x + 6 (x + ) 2 + () (x + )(x ) (x + )(x ) (x + )(x + )(x ) 6 (x ) (x + ) 2 (x ) + () (x + ) (x + )(x )(x + ) (x + )(x ) 6 (x ) +(x + ) (x + ) 2 (x ) x2 6x +6+x + (x + ) 2 (x ) x2 x +6 (x + ) 2 (x ) (x 2)(x 3) (x + ) 2 (x ) Mking Mth Possible 4 of 6 c S diyy Hendrickson

5 Rtionl Expressions Recll tht to multiply two frctions, we simply multiply the numertors together nd write the product on top. Then multiply the denomintors together nd write the product on the bottom. Algebriclly: b c d c bd Division, on the other hnd, is performed by multiplying the frction in the numertor by the reciprocl of the frction in the denomintor. Note tht we require b 6 0,c6 0 nd d 6 0.Why? Algebriclly, we hve: Consider the following exercise: b c d b c d b d c d bc x 2 +7x + 0 x +3 x +3 x x 2 2 Generl Strtegy Begin by fctoring the expressions in the numertors nd denomintors becuse it will mke it esier to simplify/reduce the frction t the end! Be sure to stte the restrictions on the frction tht you re dividing by, when multiplying by the reciprocl nd whenever you cncel expressions! x 2 +7x + 0 x +3 x +3 x x 2 2 (x + 2)(x + ) x +3 x +3 x (x + )(x ) x +3 (x + )(x ) (x + 2)(x + )(x) x +3 provided x 6 3, ± (x + 3)(x )(x + ) x(x + 3)(x + 2)(x + ) (x + 3)(x )(x + ) x(x + 3)(x + 2)(x + ) provided x 6 3, x, provided x 6 3, ± x(x + 2) Note: Restrictions my lso include x 6 0, 2. However, this is not required since the expressions x nd x + 2 re still visible in the denomintor, mking these restrictions evident when looking t the finl, simplified expression. Here, it is lso importnt to note tht the only reson we were ble to cncel the bove expressions is becuse we first cknowledged the restrictions: x nd x 6 0. Consequently, x 6 3, x +3 nd thus, x +3 x +. Furthermore, we know tht multiplying by does not chnge the vlue x + of our finl expression! Mking Mth Possible of 6 c S diyy Hendrickson

6 Rtionl Expressions Exercise: Simplify the following expression: x + x 4 x x x 2 6 x Mking Mth Possible 6 of 6 c S diyy Hendrickson