{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 274 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times " 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 53 "High School Modul es > Precalculus by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " \+ " }{TEXT 256 18 "Synthetic Division" }}{PARA 0 "" 0 "" {TEXT -1 61 " \nDeveloping the method of synthetic division of polynomials.\n" }} {PARA 0 "" 0 "" {TEXT 258 153 "[Directions : Execute the Code Resource section first. Although there will be no output immediately, these de finitions are used later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 273 8 "0. Cod e" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "PolyDivide := proc(S,T)\n loc al Q,R;\n Q := quo( S, T, x, 'R'):\n print(S/T = Q + R/T);\n \+ end proc:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1060 "LongDiv := \+ proc( P, r)\n local A, C, d, i,j,k,cols,rows,q,L2,dg, Q,R;\n d := degree( P); cols := d + 3; rows := 3*d + 3; \n for k from \+ 0 to d do C||k := coeff(P, x, k) ; od;\n A := array( [seq( [ \+ seq(` `, j = 1..cols ) ], i = 1..rows) ]);\n\n A[3,1] := x-r;\n \+ for k from 0 to d do A[3,d - k + 3] := C||k*x^k ; od;\n for k f rom 0 to d do A[2,d - k + 3] := `__`; od;\n A[k,2] := `|`; \n\n \+ for k from 1 to d do \n q := simplify(A[3*k,2+k]/x); A[1,2+k] := q;\n A[1+3*k,1] := A[3,1]* q; A[1+3*k,2] := ` = `;\n L 2 := expand(A[3,1]*q); dg := degree(L2);\n A[1+3*k,2+k] := coef f(L2, x, dg) *x^dg; \n A[1+3*k,3+k] := coeff(L2, x, dg-1) *x^ (dg-1);\n if (k > 1) then A[3*k,3+k] := A[3,3+k]; fi;\n for j from k+2 to k+3 do A[2+3*k,j] := `__`; od;\n A[3 + 3*k,2+k] \+ := A[3*k,2+k] - A[1+3*k,2+k]; \n A[3 + 3*k,3+k] := A[3*k,3+k] - A[1+3*k,3+k]; \n od;\n print(A);\n Q := quo( P , (x-r), x, \+ 'R'):\n print(` `);print(` `);\n print(P/(x+r) = Q + R/(x+r));pr int(` `);\n end proc:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 542 "SynDiv := proc( P, r)\n local A, C, d, i,j,k;\n d := degr ee( P);\n for k from 0 to d do C||k := coeff(P, x, k) ; od;\n \+ A := array( [seq( [ seq(` `, j = 0..(d+2) ) ], i = 1..4) ]);\n \+ A[1,1] := r;\n for k from 0 to d do A[1,d - k + 3] := C||k ; od; \n for k from 0 to d do A[2,d - k + 3] := 0; od;\n for k from \+ 0 to d do A[3,d - k + 3] := `__`; od;\n for k from 0 to d do \n \+ A[4,k + 3] := A[1,k + 3] + A[2,k + 3];\n if(k " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 272 32 "1. Long Division of Polynomia ls" }}{PARA 0 "" 0 "" {TEXT -1 53 "\nWe know how to divide polynomials by long division.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Long Div( x^2 + 5*x + 2, -3 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "LongDiv( x^3 - 4*x^2 + 3*x + 7, + 2 );" }}}{PARA 0 "" 0 "" {TEXT -1 43 "\nNow Lets find a slicker way of doing this!" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 270 36 "2. The Method of Synthetic Di vision" }}{PARA 0 "" 0 "" {TEXT -1 205 "\nIf we abstract the long divi sion process as much as possible we end up with what is called synthet ic division. Lets look at an example and then we'll see how it can be \+ converted to synthetic division.\n \n\n" }{TEXT 263 24 " Exampl e 2.1: " }{TEXT -1 1 " " }{XPPEDIT 18 0 "(x^2+5*x+2)/(x+3);" "6#*&, (*$%\"xG\"\"#\"\"\"*&\"\"&F(F&F(F(F'F(F(,&F&F(\"\"$F(!\"\"" }{TEXT -1 3 "\n\n " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "LongDiv( x^2 + 5 *x + 2, -3 );" }}}{PARA 0 "" 0 "" {TEXT -1 270 "\nTo divide synthetica lly .... \n we set it up in this way....create a table with the coefficients of the numerator along the top (right) and\n the \+ root of the binomial divisor in the upper left corner.\n\n then we follow these steps :\n\n " }{TEXT 259 10 "BRING DO WN" }{TEXT -1 86 " : bring down the leading coefficient ... all the wa y to the bottom\n\n " }{TEXT 260 11 "MULTIPLY UP" } {TEXT -1 88 " : multiply the root (-3) by this number, and bring it up to 2nd row in the next column\n" }}{PARA 0 "" 0 "" {TEXT -1 102 " \+ we repeat these steps until we have finished the table and there is nothing left to bring down\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "SynDiv( x^2 + 5*x + 2, -3 );" }}}{PARA 0 "" 0 "" {TEXT -1 601 " \nThere are many things to notice about the bottom row of this table. \+ If you compare it with the quotient and remainder we computed just abo ve this, you'll see that the last entry in the bottom row is the remai nder, and the other members of that row are the coefficients of the qu otient. Of course the quotient has degree one less than the numerator \+ since we are dividing by a binomial of degree one. \n\nThe advantages \+ of this system are that its fast, easy, and less prone to errors. We m erely need to read the answer off of the resulting last row to perform divisions.\n\nLets look at more examples.\n \n\n" }{TEXT 262 24 " \+ Example 2.2: " }{TEXT -1 1 " " }{XPPEDIT 18 0 "(x^3 + 11*x^2 - \+ 13*x + 8) /(x+5)" "6#*&,**$%\"xG\"\"$\"\"\"*&\"#6F(*$F&\"\"#F(F(*&\"#8 F(F&F(!\"\"\"\")F(F(,&F&F(\"\"&F(F/" }{TEXT -1 6 "\n\n " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "N := x^3 + 11*x^2 - 13*x + 8; \nLon gDiv( N, -5 );\nSynDiv( N, -5 );" }}}{PARA 0 "" 0 "" {TEXT -1 4 "\n \+ " }}{PARA 0 "" 0 "" {TEXT 261 27 " \n\n\n Example 2.3: " } {TEXT -1 1 " " }{XPPEDIT 18 0 "(4*x^3+x^2-20*x+3)/(x-4);" "6#*&,**&\" \"%\"\"\"*$%\"xG\"\"$F'F'*$F)\"\"#F'*&\"#?F'F)F'!\"\"F*F'F',&F)F'F&F/F /" }{TEXT -1 3 "\n\n " }{TEXT 265 1 " " }{TEXT -1 59 "\nWhen we divide by a binomial like (x - 4), the root is +4." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "N := 4*x^3 + x^2 - 20*x + 3; \nLongDiv( N , +4 ); \nSynDiv( N, +4 );" }}}{PARA 0 "" 0 "" {TEXT 266 26 "\n\n\n Examp le 2.4 : " }{TEXT -1 1 " " }{XPPEDIT 18 0 "(7*x^3-9*x+11)/(x+2);" " 6#*&,(*&\"\"(\"\"\"*$%\"xG\"\"$F'F'*&\"\"*F'F)F'!\"\"\"#6F'F',&F)F'\" \"#F'F-" }{TEXT -1 3 "\n\n " }{TEXT 264 1 " " }{TEXT -1 131 "\n\nJust \+ as with long division, we need to remember to place zero terms as plac eholders for any missing terms. That much is the same." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "N := 7*x^3 - 9*x + 11; \nLongDiv( N , -2 );\nSynDiv( N, -2 );" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT 271 49 " 3. Synthetic Division In More Complicated Cases" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "\nThis one is a little trickier ...." } {TEXT 268 27 " \n\n Example 3.1 : " }{TEXT -1 1 " " } {XPPEDIT 18 0 "(x^3+6*x^2+5*x+13)/(4*x-3);" "6#*&,**$%\"xG\"\"$\"\"\"* &\"\"'F(*$F&\"\"#F(F(*&\"\"&F(F&F(F(\"#8F(F(,&*&\"\"%F(F&F(F(F'!\"\"F3 " }{TEXT -1 3 "\n\n " }{TEXT 269 1 " " }{TEXT -1 218 "\n\nWhen dividin g by a binomial with a coefficient in front of the x, we can use this \+ trick ... factor that coefficient out. The quotient will be divided by that additional factor : \n " } {XPPEDIT 18 0 "(x^3+6*x^2+5*x+13)/(4*x-3) = (x^3+6*x^2+5*x+13)/(4*(x-3 /4));" "6#/*&,**$%\"xG\"\"$\"\"\"*&\"\"'F)*$F'\"\"#F)F)*&\"\"&F)F'F)F) \"#8F)F),&*&\"\"%F)F'F)F)F(!\"\"F4*&,**$F'F(F)*&F+F)*$F'F-F)F)*&F/F)F' F)F)F0F)F)*&F3F),&F'F)*&F(F)F3F4F4F)F4" }{TEXT -1 3 "\n \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "print(4*x - 3 = 4,` * (`,x - 3/4,`) `);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "N := 4*x^3 - 9*x + 1 1; \nSynDiv( N, +3/4 );" }}}{PARA 0 "" 0 "" {TEXT -1 49 "\nThis is not the answer. We need to divide by 4.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "QQ := 4*x^2 + 3*x - 27/4; RR:= 95/16;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "QQ/4 + (RR)/(4*x -3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "PolyDivide( N, 4*x - 3);" }}}{PARA 0 "" 0 "" {TEXT -1 93 "Notice that we only divide the quotient by 4. T he remainder is undivided, so it stays as is.\n" }{TEXT 275 27 "\n\n\n \n Example 3.2 : " }{TEXT -1 1 " " }{XPPEDIT 18 0 "(x^3-4*x^2- 122)/(7-x);" "6#*&,(*$%\"xG\"\"$\"\"\"*&\"\"%F(*$F&\"\"#F(!\"\"\"$A\"F -F(,&\"\"(F(F&F-F-" }{TEXT -1 3 "\n\n " }{TEXT 276 1 " " }}{PARA 0 "" 0 "" {TEXT -1 161 "We use a similar trick on this problem - factor -1 \+ out of the divisor to use synthetic division, then divide the quotient we get from synthetic division by -1. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "print(7 - x = -1,` * (`,x - 7,`)`);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "N := x^3 -4*x^2 -122; \nSynDiv( N, +7 ); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "QQ := x^2 + 3*x - 21; R R:= 25;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "QQ/(-1) + (RR)/( 7-x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "PolyDivide( N, 7 - x);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n\n" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT 274 18 " 4. Finding Roots" }}{PARA 0 "" 0 "" {TEXT -1 190 "\nWh enever we divide a polynomial by a binomial and the remainder is zero, that means the binomial divides into the polynomial. And thus the roo t of the binomial is a root of the polynomial.\n" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 37 "SynDiv( x^4 + x^3 + x^2 + x + 1, -1);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "SynDiv( 168*x^3+731*x^2-862 *x+96, 6/7);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "SynDiv( 6 *x^4-19*x^3-106*x^2+21*x+18, -3 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "SynDiv( 6*x^4-19*x^3-106*x^2+21*x+18, -1/3 );" }}}} {PARA 0 "" 0 "" {TEXT 267 36 "\n \251 2002 Waterloo Maple Inc \+ " }{TEXT -1 0 "" }}}{MARK "2 0" 1 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }