Factoring Polynomials

Transcription

1 Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles rised to some non-negtive integer exponents. A polynomil with only one term is clled monomil, with two terms is clled binomil nd with three terms is clled trinomil. A polynomil in vrible x ( single-vrible polynomil) is of the form: p(x) = n x n + n x n + n x n x + x + 0 where ll of the coefficients ( n, n, n,... +,, 0 ) re rel numbers nd ll of the exponents (n, n, n,...,,, ) re non-negtive integers. The term 0 is clled the constnt term. The degree of term of polynomil is the sum of the exponents of the vribles of tht term. The degree of polynomil is the lrgest degree of ny of the individul terms of tht polynomil. The root of polynomil p(x) is number,, such tht p() = 0. The root or x-intercept or zero of the corresponding polynomil eqution p(x) = 0 is the number,, tht mkes p() = 0. If is root of polynomil p(x), then p(x) = (x )(q(x) for some polynomil q(x). The process of identifying some or ll such fctors (x ) is clled fctoring the polynomil. There re mny different types of fctoring techniques tht cn be pplied, none of which work for ll polynomils, so determining which fctoring method to be pplied requires some skill nd persistence. The Fundmentl Theorem of Algebr sttes: Every polynomil eqution hving complex coefficients nd degree n hs t lest one complex root. (On lighter note, the Frivolous Theorem of Arithmetic sttes: Almost ll nturl numbers re very, very, very lrge.) This cn be written symboliclly s: p(x) = n x n + n x n + n x n x + x + 0 = n (x z n ) (x z n ) (x z n )... (x z ) (x z ) n = n ( x z i ) (clled stndrd form) i= The theorem cn lso be tken to men tht ny polynomil with rel coefficients hs t lest one root, which cn be rel or complex. It cn be extended to imply tht ny polynomil with rel coefficients cn be cn be fctored into the product of liner or qudrtic roots (which is pplicble to secondry school mthemtics). For ll remining definitions nd exmples, we shll ssume tht ll coefficients of the polynomils nd the polynomil equtions (or functions) re rel numbers, unless otherwise stted. A trinomil which hs leding coefficient of is clled monic qudrtic trinomil (e.g. x + 7x +) nd those which hve the leding coefficient (i.e. in x + bx + c) not equl to is clled nonmonic qudrtic trinomil. This distinction usully requires different pproch when fctoring. The term of the polynomil (in one vrible) with the highest exponent is clled the leding term or the dominnt term. The polynomil lso derives is nme (in terms of its degree) from this leding term. Polynomils re clssified s liner (for x + b), qudrtic (x + bx + c), cubic (x + bx + cx + d), qurtic (x 4 + bx + cx + dx + e), etc., for leding term exponents of,,, 4, etc. A polynomil of the form (x ) n is sid be degenerte (or to hve degenerte roots) s ll roots re equl to.

2 A polynomil of degree n hs exctly n roots. These roots re not ll necessrily distinct. If one root r, of polynomil occurs just once, then it is clled simple root. If root r occurs exctly m times, where m >, then it is clled root of multiplicity m (or n m-fold root). If m =, r is clled double root, if m =, it is clled triple root; nd so on. If polynomil with rtionl coefficients hs the irrtionl root + numbers) then its irrtionl conjugte, b, is lso root. b (where nd b re rtionl If polynomil with rel coefficients hs the imginry root + numbers) then its imginry conjugte, b, is lso root. b (where nd b re rel A polynomil hs 0 s root if nd only if the constnt term of the eqution is zero. If the leding coefficient of polynomil is, then ll of its rtionl roots re integers. If rtionl frction q p, expressed in lowest terms, is root of polynomil (or polynomil eqution) then p is divisor of the constnt term 0 nd q is divisor of the leding coefficient, n. Notice tht the converse of this sttement is not true. For exmple, we cn only sy tht is possible root of the polynomil 9x 4 5x + 8x + 4 (since is fctor of 4 nd is fctor of 9). Unfortuntely, this cn necessitte lengthy process of exmining mny potentil roots of polynomil before one or more is ctully identified. Once we hve found one root of polynomil, it usully becomes esier to identify its remining roots. As well, there re number of useful hints nd strtegies tht cn speed up this process (but you will hve to ttend the session on Sturdy to lern these). One such rule (tht will not be delt with on Sturdy) is briefly described here. It is clled Descrtes Rule of Signs for Polynomil Roots. It first requires the definition of the vrition of sign of polynomil. When polynomil is rrnged in descending order of powers of the vrible, if two successive terms differ in sign, the polynomil is sid to hve vrition of sign. For exmple, the polynomil 4x 5 x + 5x + x 4 hs three vritions of sign. Descrtes Rule sttes tht the number of positive roots of polynomil is equl either to the number of vritions of sign or to tht number diminished by n even number. (It is not ll tht helpful for fctoring polynomils t the high school level is it!) Ech ttempt mde to confirm tht possible root (x ) for polynomil p(x) relly is root (using either long division or synthetic division) is clled tril. If it turns out tht (x ) is root of p(x) then it is clled successful tril, nd it is clled flse tril if it is unsuccessful. Once you hve identified one root of the polynomil eqution p(x) = 0, nd hve used long or synthetic division to find q(x), the remining fctor of p(x) (i.e. p(x) = (x )q(x) = 0), then the resulting eqution, q(x) = 0 is clled the depressed eqution for p(x) = 0. The first few expnsions of the binomil theorem (x + y) n re often required in working with polynomils, nd lso led to the definitions of certin types of fctoring questions. These re: (x + y) = x + xy + y nd (x + y) = x + x y + xy + y nd (x + y) 4 = x 4 + 4x y + 6x y + 4xy + y 4 These led to the definitions or formuls for fctoring perfect squre trinomils nd cn lso extend to fctoring the difference of squres nd the sum nd difference of cubes. We hve: x y = (x y)(x + y) nd x y = (x y)(x + xy + y ) nd x + y = (x y)(x xy + y )

3 Some of the different methods used to fctor polynomils likely include: Grde 9 0 T Common fctor Common fctor Common fctor Common fctor e Difference of squres Difference of squres Difference of squres c Perfect squre trinomils Perfect squre trinomils Perfect squre trinomils h Simple trinomils Grouping Sum of cubes n Complex trinomils Simple trinomils Difference of cubes i Complex trinomils Grouping q Simple trinomils u Complex trinomils e The Fctor Theorem The bsic skills in mthemtics needed to pply the techniques include strong working knowledge of: The multipliction nd division fcts up to (or to 5 5 or even to 0 0???) Perfect squres up to (or up to 5 or even up to 0 ) Perfect cubes up to 0 Here is drill sheet tht cn be used for prcticing the rpid fctoring of monic qudrtic trinomils:.) (x )(x ).) (x )(x ).) (x )(x ) 4.) (x )(x ) 5.) (x )(x ) 6.) (x )(x ) 7.) (x )(x ) 8.) (x )(x ) 9.) (x )(x ) 0.) (x )(x ) You cn work out the following questions hed of the session if you like but it is certinly not necessry to do so. Note tht there is not enough spce on this pge to fully fctor the longer types..) Fctor the following trinomils: ) x + x + 4 Sign Pttern (designted s simple bove) Question Answer b) x x c) x + x 4 + } one + nd } one sign d) x x 4 (lrgest number hs sign of the x-term).) Fctor the following trinomils: ) 5x + 6x + (designted s complex bove) b) 4x 9x + c) 6x + x 5 d) 6x 40x + 5 e) 6x 58x + 5

4 .) Fctor the following trinomils: ) 5x y 5 + 5x 4 y + 5x y (common fctoring required) b) 4x(7m 5) 8y(5 7m) c) πr + 4πrh x x x x d) e + ( x ) e e) 5x 4y 7 9z + 4y y 5 f) ( ) ( ) 6 x 4 ( 4) ( ) 6 x + x + x x + x 4.) Fctor the following trinomils: ) 5x 69y (difference of squres or cubes) b) Ω (v ) c) 4x 0x + 5 9x + 4x 49 d) x 6 y 6 e) 5x + 4 (Use difference of squres) Evlute: ) Divide: x 4 5x + 6x 4 by x using ) forml long division or b) synthetic division (Horner s Method for evluting function in x, given x = ) ) Fctor the following trinomils: ) x x 0x + 4 (fctor theorem my be required) b) x + 9x + x + 5 c) x + 0x 8x 40 d) 6x 40x 5 7.) Solve for x: ) x = x + b) 8x 4 8x x + = 0 (Do you recognize the significnce of these two equtions?)

5 Note tht φ = + 5 (See the ttched Excel sheet on the Golden Men.) The Qudrtic Formul: x = b ± b 4c The generl formul for solving the cubic eqution x + bx + cx + d = 0 is given by: 4 { b + [ b + 9bc 7d + b c + 8 d 54 bcd + b d + ] x = c 4 [ b + 9bc 7d b c + 8 d 54 bcd + b d + ] } + c A Closer Look t The Difference of Squres The following re in incresing order of difficulty or complexity (s you go down, then cross) x 6 x 4 5 (x 5) (y + ) 4x + x + 9 y 6y 9 9x 6 x 6 5y 0 9(x 5) 6(y + ) 6x + 40x + 5 9x 48y 64 9x + 6 x 4 8 x + 0x y 4x + 6z + 5y z 9 + 0xy 9x 6y (x + 5) 6 9y x + 6x 9 x 4 + 4x + 6 4x 54 6 (x 5)

6 Some Selected Number Squring Ptterns Squring Numbers Ending In 5 Squring Numbers in the 50 s Squring Numbers in the 00 s Squring Numbers in the 90 s 5 = 5 50 = = = = 65 5 = 60 0 = = = 5 5 = = = = 5 = = 96 = = 54 = 05 = 95 = 65 = 55 = 06 = 94 = 75 = 56 = 07 = 9 = 85 = 57 = 08 = 9 = 95 = 58 = 09 = 9 = 995 = 59 = 4 = 90 = Why the First Two Ptterns Work.) e.g. 57 = Let the generl term be 50 + n, n ε I + Now (50 + n) = 50(50 + n) + n(50 + n) = n + 50n + n = n + n = 00(5 + n) + n.) e.g. 65 = 00 (6 7) + 5 Let the generl number be 0n + 5, n ε I + Now (0n + 5) = 0n(0n + 5) + 5(0n + 5) = 00n + 50n + 50n + 5 = 00n + 00n +5 = 00(n + n) + 5 = 00(n)(n+) + 5

7 Divisibility Tests Dividing by Look t the ones digit. If it is divisible by, then the number is s well. Exmple: is divisible by four lso, becuse 6 is divisible by two. Dividing by Add up the digits: if the sum is divisible by three, then the number is s well. Exmples: : the digits dd to 6 so the whole number is divisible by three The digits dd up to 57, nd 5 +7 =, so the originl number is divisible by three. Dividing by 4 Look t the lst two digits (the tens nd ones digits). If they re divisible by 4, the number is s well. Exmple: is divisible by four lso, becuse is divisible by four. Dividing by 5 If the ones digit is five or zero, then the number is divisible by 5. Dividing by 6 Check the rule for divisibility by nd by. If the number is divisible by both nd, it is divisible by 6 s well. Dividing by 7 To find out if number is divisible by seven, tke the ones digit, double it, nd subtrct it from the rest of the number. Exmple: If you hd 0, you would double the ones digit to get six, nd subtrct tht from 0 to get 4. If you get n nswer divisible by 7 (including zero), then the originl number is divisible by seven. If you don't know the new number's divisibility, you cn pply the rule gin. Here is nother method for divisibility by seven: Divide off the number into groups of digits (from the right). You my be left with group of one or two digits on the left end. Then find the sum of the lternte groups of these new -digit numbers. If the difference between the two sums is divisible by 7, then so ws the originl number.

8 Exmple: For the number divide up the number into these groups of digits Now dd up the lternte groups: = 5 nd = 67 Find the difference between theses new sums: 5 67 = 64 Since 64 is evenly divisible by 7 (5 times) then lso ws divisible by 7. Dividing by 8 Check the lst three digits (the hundreds, tens nd ones digits). If the lst three digits of number re divisible by 8, then so is the whole number. Exmple: 888 is divisible by 8; 886 isn't. Dividing by 9 Add the digits. If they re divisible by nine, then the number is s well. This holds for ny power of three. Dividing by 0 If the units digit is 0, then the number is divisible by 0. Dividing by Tke ny number, such s Add the first, third, fifth, seventh,.., digits = Add the second, fourth, sixth, eighth,.., digits = If the difference, including 0, is divisible by, then so is the number. - = 0 so is evenly divisible by. Dividing by Check for divisibility by nd 4. Dividing by Delete the ones digit from the given number. Then subtrct nine times the deleted digit from the remining number. If wht is left is divisible by, then so is the originl number. Exmple: For the number 4667, delete lst 7 nd multiply it by 9 (7 9 = 6). Now = 40. Since 40 is divisible by ( times) so too is Another method for divisibility tests for nd for involves exctly the sme process s the second method given for divisibility by 7. Divide the number into groups of three nd find the sum of lternting groups. If the difference of these sums is divisible by (or by ) then the originl number ws lso divisible by (or ).

9 Let x = t u (where t > u) Consider squre ABCD with side of length t (shown t right) Let u be the length of the side of smller squre (shded in blue in the digrm t right. Thus the pink squre (shown t right) hs sides of length t u. We lso hve two rectngles (shown in An Alternte Proof for The Qudrtic Formul t u A t u u B t u u t u t u u yellow), ech with length t u D t u u C nd width u. t The re of lrge squre (with side of length t) equls the sum of the res of the two smller (in pink nd blue) squres nd the two (yellow) rectngles. Thus we hve: t = (t u) + u(t u) + u which gives: t u = (t u) + u(t u) Now let m = u nd n = t u Hence, u = m nd thus t u = Therefore x = t u = m ± + 4 t m = n Therefore t m + n 4 n m 4 m = ± m + n = Since t u = (t u) + u(t u) we get (t u) = t u u(t u) = which gives m t = ± n m ± m + 4n or x = n mx which gives us x + mx n = 0 nd the solution for x in this eqution is: x = m ± m + 4n Of course, x + mx n = 0 cn be expressed s x + bx + c = 0 if we let This would mke the solution for x + bx + c = 0 result in x = b ± b 4c = b ± b 4c b ± = b 4c m = b nd n = c b ± b 4c x = which gives

10 Hints to use to identify possible roots when using the Fctor Theorem (In the exmples below, we men tht if (x ) is fctor then is root).) For the polynomil n x n + n x n + n x n + x + 0 (or the corresponding polynomil eqution): ) If p is divisor of the constnt term 0 then ± p is possible divisor of the polynomil, nd b) If q is divisor of the leding coefficient, n then qx ± p is possible fctor of the polynomil For exmple, the polynomil x + 6x 8x 0 could hve roots ±, ±, ±5, ±0 ±, ±, ± 5 nd ± 0. (You must still test one or more of these roots until you obtin successful tril.).) If the signs lternte from term to term in the polynomil (including the signs of terms with coefficient zero), then t lest one root is positive. Exmples of this type of polynomil include: 5x 9x + 7x 5 nd x 4 + 6x 8x + 5 (note tht it must be written: x 4 0x + 6x 8x + 5).) If the signs of every term of the polynomil re positive, then ll of the rtionl roots of the polynomil re negtive. e.g. ll of the rtionl roots of x 4 + 5x + 7x + 8x + 4 re negtive 4.) If the bsolute vlue of the lrgest coefficient of ny term of the polynomil is lrger thn the sum of the bsolute vlues of ech of the remining terms, then neither nor is root of tht polynomil. e.g. in the polynomil x +6x 5x 4, neither nor cn be root (becuse 6 > ) 5.) If the sum of the coefficients of the terms of polynomil re zero, then is root of tht polynomil. e.g. is root of x 4 9x + 4x + 8x 4 becuse = 0 6.) This one is little more difficult to see t first glnce, nd it relly only works becuse the text book uthors re mking up questions tht cn be fctored. If you re ble to mke the sum of the coefficients of ll terms of the polynomil zero by chnging one or more signs, then is root of the polynomil. Exmples of this type re: x 4 + 5x 7x + 9x + 4 (becuse equls 0) nd x 4 x + 8x + x + 5 (becuse = 0). 7.) Descrtes Rule of Signs for Polynomil Roots. This first requires the definition of the vrition of sign of polynomil. When polynomil is rrnged in descending order of powers of the vrible, if two successive terms differ in sign, the polynomil is sid to hve vrition of sign. For exmple, the polynomil 4x 5 x + 5x + x 4 hs three vritions of sign. Descrtes Rule sttes tht the number of positive roots of polynomil is equl either to the number of vritions of sign or to tht number diminished by n even number. Be wre tht it is not necessry to insert missing terms with zero coefficient (such s 0x 4 in the exmple bove) when using this rule, but it is necessry when using the rule in point (.) given bove. The suggested method for testing potentil roots in the quickest possible fshion is to use synthetic division. This method not only distinguishes between flse nd successful trils fster thn cn be obtined by using clcultor, but it lso identifies the depressed fctor (i.e. the other fctor).