{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 "" -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 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 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 }{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 "" -1 281 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Norma l" -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" }{TEXT 281 1 "\n" }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 256 18 "Rational Root Test" }}{PARA 0 " " 0 "" {TEXT -1 68 "\nAn exploration of the rational root test for fac toring polynomials\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : E xecute the Code Resource section first. Although there will be no outp ut immediately, these definitions are used later in 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 9 "r estart; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 287 "RatDiv := proc( L, C)\n local n,m, i, j;\nprint(`The possible rational binomial divisors are :`);\n n := numtheory[divisors](L);\nm := numtheory[divisors](C);\nfor i from 1 to nops(n) do\n for j from 1 to nops(m) do\n print( n[i]*x + m [j],` ` , n[i]*x - m[j] );\nod; od;\nend proc:\n\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 541 "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 1207 "RatRootTest := proc( P)\n local A, C, d, i,j,k ,n,m,row,rows;\n d := degree( P);\n for k from 0 to d do C||k \+ := coeff(P, x, k) ; od;\n n := numtheory[divisors](C||d);\n m : = numtheory[divisors](C||0);\n rows := 2*nops(n)*nops(m) + 2;\n \+ A := array( [seq( [ seq(` `, j = 0..(d+3) ) ], i = 1..rows) ]);\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] := `__`; od;\n \n for i from 1 to nop s(n) do\n for j from 1 to nops(m) do\n row := j + nops(m)* (i-1) + 2;\n A[row,1] := m[j]/n[i];\n A[row,2] := `|`;\n A[row,3] := C||d;\n for k from 1 to d do \n \+ A[row,k+3] := A[1,k+3] + A[row,1]*A[row,k+2];\n od;\n \+ if( A[row, d+3] = 0) then A[row, d+4] := `Root!`; fi;\n od: od:\n\n for i from 1 to nops(n) do\n for j from 1 to nops(m) do\n \+ row := j + nops(m)*(i-1) + 2 + nops(n)*nops(m);\n A[row,1] := -m[j]/n[i];\n A[row,2] := `|`;\n A[row,3] := C||d;\n \+ for k from 1 to d do \n A[row,k+3] := A[1,k+3] + A[r ow,1]*A[row,k+2];\n od;\n if( A[row, d+3] = 0) then A[ro w, d+4] := `Root!`; fi;\n od: od:\n\n print(A);\n end proc:\n \n\n\n" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 261 67 "1. T he Relationship Between Polynomials Factors and Expanded Result" }} {PARA 0 "" 0 "" {TEXT -1 280 "\nPlease take a look at these examples f or a moment. Look at the leading coefficients of each binomial and how they compare to the leading coefficient of the expanded polynomial. A lso look at the constant terms and how they compare to the constant te rm of the expanded polynomial.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "(2*x + 5)*(x + 3)*(4*x -9): % = expand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "(100*x + 2)*(10*x + 8)*(1000*x - 4): % = ex pand(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "(a[1]*x + b[1]) *(a[2]*x + b[2])*(a[3]*x + b[3]); sort(expand(%));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "(a[1]*x + b[1])*(a[2]*x + b[2])*(a[3]*x + b [3])*(a[4]*x + b[4]); \nsort(expand(%));" }}}{PARA 0 "" 0 "" {TEXT -1 361 "\n\nThe interesting thing you might notice is that the leading te rm of the result comes purely from the leading terms of the factors, a nd the constant term of the result comes purely from the constant term s of the factors. In fact, the leading term is the product of the lead ing terms of the factors, and constant is the product of the constant s of the factors!\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 262 36 "2. The Concept of the Rational Roots" }}{PARA 0 "" 0 "" {TEXT -1 247 "\nNow we turn this thinking ar ound. If we can see only the resulting polynomial, and we assume that \+ it has rational factors, then those binomial factors must have coeffic ients which are factors of the resulting terms. In particular, if a bi nomial, " }{TEXT 267 6 "ax + b" }{TEXT -1 23 ", divides a polynomial \+ " }{TEXT 269 1 "P" }{TEXT -1 10 "(x), then " }{TEXT 266 1 "a" }{TEXT -1 50 " must be a factor of the leading coefficient, and " }{TEXT 268 1 "b" }{TEXT -1 40 " must be a factor of the constant term.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "P := 12*x^5 + 17*x^3 + 35;" }}}{PARA 0 "" 0 "" {TEXT -1 52 "\nThe divisors of the leading coeffici ent, 12, are :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "numtheor y[divisors](12);" }}}{PARA 0 "" 0 "" {TEXT -1 46 "\nThe divisors of th e constant term, 35, are :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "numtheory[divisors](35);" }}}{PARA 0 "" 0 "" {TEXT -1 48 "\nIf we \+ form every possible binomial of the form " }{TEXT 270 6 "ax + b" } {TEXT -1 8 ", where " }{TEXT 272 1 "a" }{TEXT -1 28 " is from the firs t list and " }{TEXT 271 1 "b" }{TEXT -1 33 " is from the second, we ge t this." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "RatDiv( 12, 35); " }}}{PARA 0 "" 0 "" {TEXT -1 419 "\nAlthough this list is long, it is exhaustive. It includes EVERY possible binomial factor of this polyno mial. If none of these divides the polynomials then it is not divisibl e by a binomial with integer coefficients. So all we have to do is to \+ check these out!\n\n\nHere are other examples. Note that it depends on how many divisors the leading coefficient and constant have.. the mor e divisors, the more possible factors." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "P := x^2 + 17*x^3 + 25;\nRatDiv( 1, 25);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "P := 81*x^2 + 17*x^3 + 1;\nRatDiv( \+ 81, 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "P := 32*x^2 + 17 *x^3 + 81;\nRatDiv( 32, 81);" }}}{PARA 0 "" 0 "" {TEXT -1 162 "\nNotic e that when the two numbers are not relatively prime, there may be som e duplications among the list of potential factors - for example, (x+5 ) and (3x + 15)." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "P := 6*x ^3+5*x^2-44*x-15 ;\nRatDiv( 6, 15);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 263 31 "3. The Method of Rational Roots" }}{PARA 0 "" 0 "" {TEXT -1 227 "\nSuppose we want \+ to factor the following polynomial. We look at its leading coefficient and constant term (a.k.a. trailing coeffficient). We find all diviors of each, then form all of the possible divisors using these numbers. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "P := 5*x^3+18*x^2+7*x- 6 ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "RatDiv( 5, 6);" }}} {PARA 0 "" 0 "" {TEXT -1 86 "\nIts not pretty, but we can perform a sy nthetic division on each possible factor. If (" }{TEXT 273 4 "ax+b" } {TEXT -1 33 ") is a factor, then we will test " }{TEXT 274 4 "-b/a" } {TEXT -1 107 " in the synthetic division. We are looking for a zero re mainder. Lets try a few of the 16 possible cases :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "SynDiv( P,+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "SynDiv( P,+1/5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "SynDiv( P,+6/5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "SynDiv( P,-3);" }}}{PARA 0 "" 0 "" {TEXT -1 320 "\nWe got lucky! We found a root... x = -3. This means that (x+3) is a fact or. \n\nNow we can continue the factoring process on the quotient - wh ich is second degree and easier to deal with. We can do the rational r oot test on this smaller polynomial,....noting that there are fewer ca ndidates for being rational factors.\n\n\n" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 14 "RatDiv( 5, 2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "SynDiv( 5*x^2 + 3*x - 2, -1);" }}}{PARA 0 "" 0 "" {TEXT -1 141 "We found another root! And the quotient is a linear term .\n\n.... or in this case, we could have simply factored the quadratic expression ....\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "factor ( 5*x^2 + 3*x - 2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "P = \+ factor(P);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 264 36 "4. The Fast Method of Rational Root s" }}{PARA 0 "" 0 "" {TEXT -1 298 "\nInstead of doing all of those ind ividually, we can speed up the process by putting the synthetic divisi ons in one chart. We write the polynomial coefficients at top, and the n each of the \"bottom lines\" on one row - hiding the former middle l ines by doing those calculations in our heads.\n\n " }{TEXT 275 13 "Example 4.1 :" }{TEXT -1 5 " \n" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "P := 5*x^2 + 3*x - 2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "RatDiv( 5, 2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "RatRootTest(P);" }}}{PARA 0 "" 0 "" {TEXT -1 35 " \nHere is an other example.\n " }}{PARA 0 "" 0 "" {TEXT 276 21 " Exam ple 4.2 :" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "P := 2*x^3+17*x^2+27*x-18;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "RatDiv(2, 18);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "RatRootTest( P );" }}}{PARA 0 "" 0 "" {TEXT -1 194 " \nThe sweet site of success! We found all four roots : 1/2, -3, -6, -3 . Wait a second... this is a third degree polynomial so there are at m ost three roots. Actually, -3 is a duplicate.\n\n " }}{PARA 0 " " 0 "" {TEXT 277 21 " Example 4.3 :" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "P := x^3+5*x^2 -18*x-72;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "RatDiv(1, 72); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "RatRootTest( P);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "\nHere is about as easy as it gets!\n " }}{PARA 0 "" 0 "" {TEXT 278 21 " Example 4.4 :" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "P := x^5+x^4-2*x^3-2*x^2+x+1;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "RatDiv(1, 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "RatRootTest( P);" }}}{PARA 0 "" 0 " " {TEXT -1 106 "\nHowever, we only found two roots in a fifth degree e quation. Lets remove these roots to see what remains." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "P/((x-1)*(x+1)): % = simplify(%);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "P2 := x^3 + x^2 - x -1;\nRa tRootTest( P2 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "P = ((x -1)^2)*(x+1)^3;" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 265 42 "5. When the Method of Rational Roots Fails" }}{PARA 0 "" 0 "" {TEXT -1 52 "\nSometimes the test fails to find any roots at all.\n" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "P := 3*x^3+5*x^2-18*x + 14; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "RatDiv(3, 14);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "RatRootTest( P);" }}}{PARA 0 "" 0 "" {TEXT -1 200 "\nEvidently, this polynomial does not have any binomials with integer coefficients, and thus no rational roots. Any \+ roots are irrational. Since all odd-degreed polynomials have at least \+ one real root.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "plot( P, x = -5..4, y = -50..120, color = blue, thickness = 2 );" }}}{PARA 0 "" 0 "" {TEXT -1 109 "\nWe can use Maple's floating-point solving func tion to find a decimal approximation to this irrational root.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "fsolve( P = 0, x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 259 36 "\n \251 2002 Waterloo Maple Inc " }}}{MARK "0 1" 39 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }