MAT 51 Wldis Projet 6: Minigols Towrds Simplifying nd Rewriting Expressions The distriutive property nd like terms You hve proly lerned in previous lsses out dding like terms ut one prolem with the wy tht we often think out like terms is tht it isn t lwys ler wht is ment y like. Relly, omining like terms is relly just pplying the distriutive property in reverse: + = ( + ) Or, more generlly (for ny numer of terms, where njust denotes the numer of terms) we ould write: 1 + + n = ( 1 + + n ) Any time we hve sequene of terms, eh of whih is produt (two things eing multiplied together), nd the seond ftor in eh produt is the SAME, then we n used the distriutive property to rewrite the expression. So, now let s work on prtiing when we n pply the distriutive property. Exmples: Whih of these hve the form of the left side of one of the distriutive properties ove? For eh one, CIRCLE, nd identify,, or 1,, n, : 5x 3x 5x 3y 8p 2 q 2 12p 2 q 2 8 5 5 5 + 2 5 2 3 + 3 2 5 + 4 The distriutive property n e pplied if we rewrite the sutrtion s dding negtive first: 5x 3x = 5x + 3x So 5 x + 3 x And = 5, = 3, = x Or we ould write 1 = 5, 2 = 3, = x The distriutive property n NOT e pplied here euse these two terms don t hve nything in ommon. I nnot find nything of the form +, even if I rewrite the sutrtion s dding negtive first. The distriutive property n e pplied if we rewrite the sutrtion s dding negtive first: 8p 2 q 2 12p 2 q 2 = 8p 2 q 2 + 12p 2 q 2 So 8 p 2 q 2 + 12 p 2 q 2 And = 8, = 12, = p 2 q 2 Or we ould write 1 = 8, 2 = 12, = p 2 q 2 The distriutive property n e pplied if we rewrite the sutrtion s dding negtive first: 8 5 5 5 + 2 5 = 8 5 + 5 5 + 2 5 So 8 5 + 5 5 + 2 5 And 1 = 8, 2 = 5, 3 = 2, = 5 Note: here we hve to use the formul with 1,, n in it, euse here we hve more thn two terms. The distriutive property n NOT e pplied here euse these two terms don t hve nything in ommon. I nnot find nything of the form +, even if I use the ommuttive property = to reorder the ftors in eh term first. Note: 2 is NOT the sme s 2, nd 3 is NOT the sme s 2. The distriutive property n e pplied if we think out how to group ll the ftors in the first term, nd then group the 5 together nd tret it s one unit: 5 + 4 = (5) + 4 So (5) + 4
3x(x 1) + 2(x 1) 5x 3x 2 Here I hve s nd s in my expression nd in the distriutive property identity, so I m going to hnge the letters in the first distriutive property like this: + = ( + ) pp + qq = (p + q)r Then p = 5, q = 4, r = Or we ould write 1 = 5, 2 = 4, = The distriutive property n e pplied: So 3x (x 1) + 2 (x 1) And = 3x, = 2, = (x 1) Or we ould write 1 = 3x, 2 = 2, = (x 1) The distriutive property n e pplied if we 1) rewrite the sutrtion s dding negtive, nd think out how to group ll the ftors in the seond term, fter rememering tht x 2 n e repled with x x: 5x 3x 2 = 5x + 3 x x = 5x + ( 3x) x So 5 x + ( 3x) x And = 5, = 3x, = x Or we ould write 1 = 5, 2 = 3x, = x Do we lwys wnt to use the distriute property in exmples like this? For eh of these exmples, we ould see how we n pply the distriutive property in mny ses to omine terms, even in mny ses tht you proly didn t previously think of s like terms. In more omplex ses like these, we my or my not wnt to pply the distriutive property in reverse it will depend on our gols with the prtiulr expression. But it is importnt to understnd tht the distriutive property CAN e used in ll of these ses. Now you try! Whih of these hve the form of the left side of one of the distriutive properties ove? For eh one: ) Stte whether or not one of the distriutive properties n e pplied; ) If it n, explin whih identities, if ny, need to e pplied first; if it n NOT, explin WHY not; ) CIRCLE ; nd d) identify,, (or 1,, n,, depending on the distriutive property tht you re using) from one of the two given distriutive properties: 1. Two terms: + = ( + ) 2. Any numer of terms: 1 + + n = ( 1 + + n ) Use the exmples ove s model. 1. 2 3 2. 2 3 3. 2 2 3
4. 4x 3 y 2 + 5x 3 y 2 5. 3 7 7 6. 4 3 3 2 + 3 7. 2xx 3y 8. 2x(2x + 1) + 3(2x + 1)
Rewriting expressions with speifi gol in mind Typilly when we rewrite expressions, it is with speifi gol in mind. We my wnt to simplify them, or we my wnt to put them into prtiulr form tht llows us to see reltionships (for exmple, when we ome to equtions, we often put them in prtiulr form in order to e le to solve them or grph them). When we hve gol in mind for rewriting n expression (or eqution), we then hve to put together mny of the things tht we hve lredy een doing so tht we pply identities to the expressions one fter nother. This requires severl steps: 1) We hve to identify our end gol (e.g. Why re we rewriting the expression/eqution? Wht does it need to look like when we re done?). 2) We hve to selet prtiulr identities tht ould e pplied to the urrent expression (either in its urrent form, or with little rewriting), nd we need to onsider eh of these possile identities s wy of rewriting the expression. If the rewriting is somewht omplex, this my require us to set some smller gols tht we need to meet long the wy to meet our lrger gol. 3) Eh time we pply n identity to the expression, we hve to sk ourselves if this hs rought us loser to our gol or not.. If not, we my need to put tht work side nd to strt over y pplying different identity to the originl eqution.. If it hs gotten us loser, then we need to think out wht identity we should pply to this new rewritten funtion in order to get even loser to our gol. 4) We keep repeting this proess until we reh our gol. As prolems grow more omplex, we might hve to pply mny, mny identities sequentilly to get to our finl result. Let s just jump in nd strt trying to simplify expressions y using the identities we hve een using so fr. To keep things simple, we will only need to use the identities on the list elow s we work to simplify the next set of prolems. x = 1x = + + = + ( + ) + = + + + ( + ) = + + = () = () = ( + ) = + ( + ) = + x n = x x n mmmm ttttt x = 1 x (wheeeeee x 0) (wheeeeee 0) d (wheeeeee, d 0) + = + (wheeeeee 0) = x 2 = x (wheeeeee x 0)
Exmples: A) Rewrite the following expression, with the im of simplifying it s muh s possile: (6x 2 + 5) (2x 2 + 6) If our gol is to simplify this, we will likely wnt to rewrite this without the prentheses so tht we n try to omine wht is in eh set of prentheses with eh other. If tht is possile, it will very likely mke the expression simpler. So, our first mini-gol is: Rewrite the expression without prentheses. If the expression ws of this form: ( + ) +, then we ould use the identity ( + ) + = + + to rewrite this expression without the first set of prentheses. But we need to rewrite ll of the sutrtion in this expression s ddition first. So this leds us to nother mini-gol, whih must ome BEFORE our gol of rewriting the expression without prentheses: Rewrite ll sutrtion in the expression s ddition. Let s put these two gols in the order tht we need to do them, nd list the identities needed to omplish eh gol: Mini-gol Identities needed 1) Rewrite ll sutrtion in the expression s ddition. = + 2) Rewrite the expression without prentheses. We will ( + ) + = + + strt with the first set of prentheses nd then fous on the seond set. Now let s try to do these two steps, nd see wht hppens: Step 1: Rewrite the sutrtion in the expression s ddition, using the identity = + = (6x 2 + 5), = (2x 2 + 6) = + ( ) (6x 2 + 5) (2x 2 + 6) = (6x 2 + 5) + (2x 2 + 6) So (6x 2 + 5) (2x 2 + 6) = (6x 2 + 5) + (2x 2 + 6) ( ) = ( ) + ( ) Step 2: Rewrite the expression without (the first set of) prentheses, using the identity ( + ) + = + +. Fousing on rewriting without the first set of prentheses: = 6x 2, = 5, = (2x 2 + 6) ( + ) + = + + ( ) (6x 2 ) = ( ) + (5) + (2x 2 + 6) = (6x 2 ) So (6x 2 + 5) + (2x 2 + 6) = 6x 2 + 5 + (2x 2 + 6) + (5) + (2x 2 + 6) Now we need to fous on rewriting the expression without the seond set of prentheses. The su-expression (2x 2 + 6) lmost looks like the left side of the identity ( + ) = +, exept tht the vrile n t stnd in for negtive sign lone the negtive sign hs to elong to numer or n expression. So now we form some new mini-gols, on our wy to the gol of rewriting the expression without prentheses: Mini-gol: Put 2x 2 + 6 into the form ( + ). We n do this if we notie tht we n pply the identity x = 1x to the expression (2x 2 + 6). So, let s do tht: First let s rewrite the identity x = 1x with nother vrile to void onfusion: q = 1q. q = (2x 2 + 6), q = 1q ( ) q q = 1 ( ) So 6x 2 + 5 + (2x 2 + 6) = 6x 2 + 5 + 1(2x 2 + 6) q (2x 2 + 6) q = 1 (2x 2 + 6)
Now, k to our first mini gol: Mini-gol: Rewrite the expression without the seond set of prentheses. We see tht we n now use the identity ( + ) = + to rewrite the seond hlf of the expression without prentheses: = 1, = 2x 2, = 6 ( + ) = + ( ) ( 1) (2x 2 ) + (6) = ( 1) (2x 2 ) + ( 1) (6) ( ) So 6x 2 + 5 + 1(2x 2 + 6) = 6x 2 + 5 + ( 1)(2x 2 ) + ( 1)(6) + ( ) = ( ) ( ) So now we hve simplified the originl expression to this equivlent expression: 6x 2 + 5 + ( 1)(2x 2 ) + ( 1)(6) How n we simplify this expression further? The most ovious step is for us to perform the multiplition in the lst few terms: 6x 2 + 5 + ( 1)(2x 2 ) + ( 1)(6) = 6x 2 + 5 + ( 1)(2x 2 ) + 6 euse ( 1)(6) = 6 6x 2 + 5 + ( 1)(2x 2 ) + 6 = 6x 2 + 5 + 2x 2 + 6 euse ( 1)(2x 2 ) = 1 2 x 2 = 2x 2 So now we hve simplified the originl expression to this equivlent expression: 6x 2 + 5 + 2x 2 + 6 Wht might help to mke this simpler? It would e good for us to look to see if we n omine ny of these terms together. For exmple, if 5 nd 6 were next to eh other, we ould dd them. Similrly, if 6x 2 nd 2x 2 were next to eh other, we ould use the identity ( + ) = + to omine them. So, our next mini-gol is: ( ) Mini-gol: Reorder the terms so tht like terms re next to eh other. I n use the identity + = + to do this: = 5, = 2x 2 + = + ( ) = ( ) (5) + ( 2x 2 ) = ( 2x 2 ) + (5) So 6x 2 + 5 + 2x 2 + 6 = 6x 2 + 2x 2 + 5 + 6 Now we n simply dd the finl two like terms: 6x 2 + 2x 2 + 5 + 6 = 6x 2 + 2x 2 + 1 euse 5 + 6 = 1 And now our finl mini-gol is: Mini-gol: Used the identity ( + ) = + to omine the first two terms. We noted tht the first two terms hve the form of the right side of the identity. = 6, = 2, = x 2 ( + ) = + ( ) (6) + ( 2) (x 2 ) = (6) (x 2 ) + ( 2) (x 2 ) ( ) = ( ) So, 6x 2 + 2x 2 + 1 = (6 + 2)x 2 + 1 euse (6 + 2)x 2 = 6x 2 + 2x 2 ( ) And finlly, we n dd the 6 nd the 2 inside the prentheses, so we go hed nd do tht: (6 + 2) x 2 + 1 = 4 x 2 + 1 euse 6 + 2 = 4 So now we hve repled the originl expression with the equivlent expression 4x 2 + 1 This is out s simple s it gets, lthough, we ould do one finl step nd rewrite the ddition of negtive s sutrtion, if we wnt: = 4x 2, = 1 = + ( ) ( ) = ( ) + ( ) (4x 2 ) (1) ( ) = (4x 2 ) + (1)
So, 4x 2 + 1 = 4x 2 1. This is definitely s simple s we ould mke this expression, so our finl expression is 4x 2 1 nd: (6x 2 + 5) (2x 2 + 6) = 4x 2 1 Now you try! We re going to strt trying to rewrite expressions with speifi gol in mind. However, first you re going to strt y working on prolems where you will e given severl minigols in order. You should try to omplete eh minigol, nd tht will help led you to the overll gol. Be sure to do the minigols in order, nd e sure to work sequentilly, pplying identities to the results of the previous step, NOT to the originl expression. In mny sts, you will first hve to use severl identities to rewrite the expression efore you n omplish the minigol. Use the previous exmple s guide to help you work through these prolems. 1. Rewrite the following expression, with the im of simplifying it s muh s possile: (2x + 3) (3x + 1) Minigol 1: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 2: Apply the identity + = ( + ) in order to omine like terms. 2. Rewrite the following expression, with the im of simplifying it s muh s possile: (2x + 3)(3x + 1) Minigol 1: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 2: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 3: Apply the identity + = ( + ) in order to omine like terms.
3. Rewrite the following expression, with the im of simplifying it s muh s possile: 3x 2 2x + 5x 2 5 Minigol 1: Apply the identity + = ( + ) in order to omine like terms. 4. Rewrite the following expression, with the im of simplifying it s muh s possile: 2xy 3 2xx (x, y 0) Minigol 1: Apply the identity = to rewrite the expression so tht the whole expression is under single rdil. Minigol 2: Rewrite ll multiplition under the rdil s produt of squres, insofr s possile. Prt A: Rewrite ll the multiplition under the rdil sign so tht numers re next to numers, nd the sme letters re next to eh other, y using the identity = s mny times s neessry.
Prt B: Use the identity = n to rewrite ll exponents s multiplition. n ttttt Prt C: Use the identity = 2 to rewrite ll multiplition s multiplition of squres (inluding numers nd vriles), wherever possile. Minigol 3: Use the identity 2 = to rewrite the expression without rdils, where possile. 5. Rewrite the following expression, with the im of simplifying it s muh s possile: 6x 2 3x 3x (x 0) Minigol 1: Use the identity + = + to rewrite this expression s two frtions, so tht you n fous on one frtion t time. Minigol 2: Use the identity = to rewrite the expression so tht something of the form p is isolted y itself. d d p
6. Rewrite the following expression, with the im of simplifying it s muh s possile: 2x 2 y 3 Minigol 1: Use the identity = d to rewrite the expression so tht something of the form p p x 4 is isolted y itself. 7. Put the following expression into x 2 + + form, where,, re rel numers: 2x(3x 6) + 5 Minigol 1: Apply the identity ( + ) = + in order to rewrite the expression without prentheses. Minigol 2: Apply the identity + = ( + ) in order to omine like terms.