{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 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 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 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 128 128 1 0 1 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 0 128 128 1 0 1 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 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 278 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 1 14 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 283 "" 0 1 0 128 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 128 0 1 0 0 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 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 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" }{TEXT 285 1 "\n" }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 256 38 "The Remainder Theorem & Factor T heorem" }}{PARA 0 "" 0 "" {TEXT -1 66 "\nExposition and application of the factor and remainder theorems.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : Execute the Code Resource section first. Although t here will be no output immediately, these definitions are used later i n this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 260 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "ccc := \+ 'color = COLOUR(RGB, .6, .5, .5), thickness = 2':" }}}{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 from 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 L2 := expand(A[3,1]*q); dg := degree(L2);\n A[1+3*k,2+k] := coeff(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));print(` `);\n end proc:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "PolyLinFact := proc(S,T) \n local Q,R;\n Q := quo( S, T, x, 'R'):\n print( expand('S'/ T) = T*(sort(Q,x) + R/T));\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 := degree( 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 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "N := 4*x^3 + x^2 - 20*x + 3; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "N := (x-3)*(x+5)*(x-4); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "PolyLinFact( (x-3)*(x+ 5)*(x-4), x - 3);\nPolyLinFact( (x-3)*(x+5)*(x-4), (x - 3)*(x-4));" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "SynDiv( N, +4 );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "SynDiv( x^4 + x^3 + x^2 + x \+ + 1, -1);" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 261 24 "1 . The Remainder Theorem" }}{PARA 0 "" 0 "" {TEXT -1 20 "\nOne version \+ of the " }{TEXT 264 17 "Remainder Theorem" }{TEXT -1 13 " is this :\n \n " }{TEXT 263 3 " " }{TEXT 268 133 "For a polynomial f(x), the fol lowing are equivalent :\n - f(r) = R\n - the remainder w hen (x-r) is divided into f(x) is R" }{TEXT 269 1 "\n" }{TEXT -1 130 " \nIt makes a connection between the remainder of a polynomial division and evaluating a polynomial. Let's look at some examples. \n\n" } {TEXT 266 13 "Example 1.1 :" }{TEXT -1 136 " Let's create a polynomial and pick a number r. Then we will evaluate both of the expressions ab ove and note that the two are the same.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f := x -> 3*x^2 - 7*x + 11; \nr := 4;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "LongDiv( f(x), r);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n \n" }{TEXT 267 13 "Example 1.2 :" }{TEXT -1 28 " Here is another exam ple. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "f := x -> x^3 + 1 0*x^2 - 100*x + 300; \nr := -1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "LongDiv( f( x), r);" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 265 13 "Example 1.3 :" }{TEXT -1 27 " Here is a harder example.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "f := x -> x^7 - x^6 + 4*x^5 - 9*x^4 + 11*x^3 -2* x^2 + 8*x^2 - 12*x + 144; \nr := -7/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Lo ngDiv( f(x), r);" }}}{PARA 0 "" 0 "" {TEXT -1 118 "\nIn all these case s, you can see that the number gotten from evaluating the polynomial i s the same as the remainder.\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 262 51 "2. Evaluating a Polynomial using Synthetic Divisi on" }}{PARA 0 "" 0 "" {TEXT -1 145 "\nOne application is to evaluate a polynomial by finding the remainder. In some cases, synthetic divisio n is \nfaster than the actual evaluation.\n\n" }{TEXT 271 13 "Example \+ 2.1 :" }{TEXT -1 48 " It might be debatable as to which is easier. \n " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "f := x -> x^4 + x^3 + x^ 2 + x + 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(-5);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "SynDiv( f(x), -5 );" }}} {PARA 0 "" 0 "" {TEXT -1 2 "\n\n" }{TEXT 270 13 "Example 2.2 :" } {TEXT -1 62 " Here is another example. Which method is easier and fas ter?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "f := x -> x^6 - 17 0*x^4 + 200*x^2 - 400*x + 11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(13);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "SynDiv( f( x), 13 );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT 283 15 "Example 2.3 : " }{TEXT -1 26 "Here is another example. \n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 117 "f := x -> x^11 + 3*x^10 - 7* x^9 + 20*x^8 + 30*x^7 + 31*x^6 \n + 34*x^5 - 110*x^3 - 61*x^2 - 58*x - 11;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(- 5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "SynDiv( f(x), -5 ); " }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 272 49 "3. Finding A Remainder by Evaluating a P olynomial" }}{PARA 0 "" 0 "" {TEXT -1 143 "\nWe can use the theorem in the other direction too. In some cases, its easier to evaluate a poly nomials than it is to do a synthetic division.\n" }{TEXT 282 13 "Examp le 3.1 :" }{TEXT -1 26 " Find the remainder when " }{XPPEDIT 18 0 "x^ 4+x^3+x^2+x+1;" "6#,,*$%\"xG\"\"%\"\"\"*$F%\"\"$F'*$F%\"\"#F'F%F'F'F' " }{TEXT -1 22 " is divided by (x+2).\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "f := x -> x^4 + x^3 + x^2 + x + 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(-2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "SynDiv( f(x), -2 );" }}}{PARA 0 "" 0 "" {TEXT -1 2 " \n\n" }{TEXT 281 13 "Example 3.2 :" }{TEXT -1 26 " Find the remainder when " }{XPPEDIT 18 0 "x^20+x^10+1;" "6#,(*$%\"xG\"#?\"\"\"*$F%\"#5F' F'F'" }{TEXT -1 117 " is divided by (x -1).\n\nThis synthetic division is a bit large, but its quite easy to simply evaluate the polynomial. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "f := x -> x^20 + x^10 \+ + 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f( 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "SynDiv( f(x), 1 );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 273 13 "Example 3.2 :" }{TEXT -1 26 " Fin d the remainder when " }{XPPEDIT 18 0 "x^100 + x^75 + x^50 + x^25 + 1 " "6#,,*$%\"xG\"$+\"\"\"\"*$F%\"#vF'*$F%\"#]F'*$F%\"#DF'F'F'" }{TEXT -1 115 " is divided by (x+1).\n\nWarning: don't try doing the syntheti c division unless you have a very large piece of paper!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "f := x -> x^100 + x^75 + x^50 + x^2 5 + 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(-1);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " } {TEXT 274 21 "4. The Factor Theorem" }}{PARA 0 "" 0 "" {TEXT -1 5 "\nT he " }{TEXT 276 14 "Factor Theorem" }{TEXT -1 45 " is a corollary of t he Remainder theorem :\n\n " }{TEXT 275 3 " " }{TEXT 279 131 "For a \+ polynomial f(x), the following are equivalent :\n - f(r) = 0\n \+ - (x-r) divides f(x)\n - r is a root of f(x)" }{TEXT -1 1 "\n" }{TEXT 280 61 " - the graph of f(x) has an x-intercept a t x = r\n " }{TEXT -1 114 "\n\nThis theorem makes a connection bet ween the roots of a polynomial, and linear factors. Let's see it in ac tion.\n\n" }{TEXT 277 13 "Example 4.1 :" }{TEXT -1 2 " \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f := x -> 6*x^3+13*x^2-4; " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "r := -2; f(r);\nr := 1/2; \+ f(r);\nr := -2/3; f(r);" }}}{PARA 0 "" 0 "" {TEXT -1 75 "This polynom ial has three roots. We would expect three linear factors ....." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "factor(f(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot( f(x), x = -3..2, y = -20..50, ccc );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "rt := solve( f( x) = 0, x);\ndisplay( pointplot( \{[rt[1],0],[rt[2],0],[rt[3],0]\}, sy mbolsize = 30, color = red),\n plot( f(x), x = -3..2, y = -20. .50, ccc ));" }}}{PARA 0 "" 0 "" {TEXT -1 4 "\n\n\n\n" }{TEXT 278 13 " Example 4.2 :" }{TEXT -1 28 " Here is another example. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "f := x -> 5*x^3+7*x^2+7*x+2; \nr := -2/5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "factor(f(x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "display( pointplot( [r,0], symbolsize = 30, color = red), \n plot( f(x), x = -2..1, y = -15..20, \+ ccc ));" }}}{EXCHG }{PARA 0 "" 0 "" {TEXT -1 153 "\nWhenever we have a root, we have a linear factor, and thus we have a way of decomposing \+ the polynomial into a product of smaller degreed polynomials.\n\n " }} }{PARA 0 "" 0 "" {TEXT 259 35 "\n \251 2002 Waterloo Maple Inc " }}}{MARK "0 1" 23 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }