advanced algeb project.docx

7/21/2019 ADVANCED ALGEB PROJECT.docx

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ProjectIn

Advanced

AlgebraDiomyka C. Damgo

IV-A



Mr. Reynaldo F.

Sumil

Polynomial Functions1. f ( x ) = x 3 -2 x 2 - 5 x + 6 has roots at x = { -2, 1, 3} . Sketch the function by re!ictin" its en!

beha#ior an! testin" the #a$ue of the function at %-#a$ues bet&een the roots.

2. 'o& any #ariations are in the function f ( x ) = x 6 -6 x +2 x 2 + 2 x - * Answer: Three. You just count how many times the sign changes.

3. 'o& any ositi#e an! ne"ati#e rea$ roots i"ht the function f ( x ) = 2 x 3 -3 x 2 - x + 1 ha#eAnswer: f ( x ) has two variations, so it may have two or zero positive roots. f ( x ) hasone variation, so it has one negative root.

!. hat are the ossib$e rationa$ roots of f ( x ) = x 3 - x 2 + 3 x - 6 se escartes/ 0u$e ofSi"ns an! the 0ationa$ 0oot heore.Answer: "escartes# $u%e o& 'igns inicates that there are either three or one positiverea% roots, an no negative rea% roots. The $ationa% $oot Theorem inicates that thepossi%e rationa% roots are *1 , *2 , *3 , an *+ . nowing that there are no negativeroots, the possi%e rationa% roots are 1 , 2 , 3 , an + .

-. in! a$$ rea$ roots of x 3 + x 2 - x - 2 .

Answer: x / 2, , 20 .

+. in! a$$ rea$ roots of x +5 x 3 - 15 x + .Answer: x / 3, 1, 3.12, 4.120

. 4 factory has starte! to ro!uce cars for 1 onths. ts onth$y ro!uction can be ae!

by the function 7 = .1(-t+15t3-166t2+138t+6), &here 7 is easure! in thousan!s



of cars an! t is the nuber of onths since the factory starte! car ro!uction. urin" &hich

onths !i! the factory ro!uce 1 thousan! cars

'o%ution: 'o%ve the e5uation 144 4.441(t!61-t31++t261374t6+444) to get 144444

t!61-t31++t261374t6+444 an 4 t!61-t31++t261374t!444. 8raphing this with a

ca%cu%ator, we can see that the rea% zeros o& this &unction are at - an 7.

7. o has starte! to co$$ecte! "ae car!s for 12 onths. 'is onth$y co$$ection can be

ae! by the function 9 = -t+13t3-3t2+13t+8, &here 9 is the nuber of "ae car!s an!

t is the nuber of onths since he starte! the co$$ection. urin" &hich onths !i! o

co$$ect "ae car!s

'o%ution: 'o%ve the e5uation 4 t!613t3!3t2613t6!7 to get 4 t!613t3!3t2613t!2.

8raphing this with a ca%cu%ator, we can see that the rea% zeros o& this &unction are at +

an .

. Sho& that a$$ rea$ roots of the e:uation $ie bet&een -

an! .

'o%ution: 9n other wors, we nee to show that ! is a %ower oun an ! is an upperoun &or rea% roots o& the given e5uation.hec;ing the <ower =oun: <ets app%y synthetic ivision with ! an see i& we geta%ternating signs:

>ote how c ! ? 4 A>" the successive signs in the ottom row o& our synthetic ivisiona%ternate. ! is a %ower oun &or the rea% roots o& this e5uation.hec;ing the @pper =oun: <ets app%y synthetic ivision with ! an see i& we get a%%positive:

>ote how c ! 4 A>" the a%% o& the signs in the ottom row o& our synthetic ivisionare positive. ! is an upper oun &or the rea% roots o& this e5uation. 'ince ! is a %oweroun an ! is an upper oun &or the rea% roots o& the e5uation, then that means a%% rea%

roots o& the e5uation %ie etween ! an !.

14. ;onsi!er 7(%) = %2 + 5% + 6. in! the <eros.'o%ution: The zero o& the po%ynomia% can e otaine y e5uating B(C) to 4C2 6 -C 6 + 4(C 6 3)(C 6 -)4C 6 3 4 (or) C 6 - 4C 3 , -



11.

12. 7(%) = %3 + 5%2 - % -2 = (%2(% + 5) (% + 5)) = (% + 5)(%2 - ) = (% + 5)(% + 2)(% - 2)'o%ution:The zero o& the po%ynomia% can e otaine y e5uating B(C) to 4B(C) 4

C3 6 -C2 !C 24 4(C2(C 6 -) ! (C 6 -)) 4(C 6 -)(C2!) 4(C 6 -)(C 6 2)(C 2) 4C 6 - 4, C 6 2 4, C 2 4C -, 2, 2

13. So$#e the o$ynoia$ (%) = %2 + % + 3

1!. rite an e:uation of the o$ynoia$ function of !e"ree 3 &hich has <eros of 1, 2 an! 3.'o%ution: >o&, for fin!in" the ori"ina$ for of o$ynoia$, u$ti$yin" a$$ these abo#e ters,&e "et(% - 1)(% - 2)(% - 3) =

%3 - 6%2 + 11% - 6 = ;onsi!er the o$ynoia$, 7(%) = %3 - 23 %2 + 12 % - 12. 4s &e cannot factorise this by "rouin",$et us $ist the ossib$e factors of 12.actors of 12 are , ?1, ?2, ?3, ?, ?5, ?6, ?8, ?1, ?12, ?15 ?2, ?2, ?3, ?, ?6, ?12,@et us $u" in the abo#e #a$ues one by one,7(1) =13 - 23 (12) + 12 (1) - 12 = 1 - 23 + 12 - 12 = 13 - 13 = (% - 1) is one of the factor of the abo#e o$ynoia$.

4$yin" synthetic !i#ision,



'ence the Auotient = :(%) = %2 - 22% + 12= %2 - 12 % - 1 % - (12 % 1) = %(% - 12) - 1(% - 12 ) B factori<in" by "rouin" C = (% -

12)(% - 1)'ence the factori<e! for of the o$ynoia$ 7(%) is,7(%) = (% - 1) (% - 12)(% - 1)he <ero of the o$ynoia$ can be obtaine! by e:uatin" 7(%) to 7(%) = (% - 1) (% - 12)(% - 1) % - 1 = , % - 12 = , % - 1 = % = 1, 12, 1The zeros o& the given cuic po%ynomia% are 1,14 an 12.

1-.

1+.

1.



17.

1.

Identity Equations24. 21. 22.

'hannon, a cainetma;er, starte out with a %oc; o& woo, an then she ho%%owe outthe center o& the %oc;. The imensions o& the %oc; an the cutout is shown e%ow.



2!.

"eterminea%geraica%%y

whether &(C)

D

3C26 !is even, o, or neither.

4>SD0Ef "rah this, &i$$ see that this is Fsyetric about the y -a%isFG in other &or!s,

&hate#er the "rah is !oin" on one si!e of the y -a%is is irrore! on the other si!eEhis irrorin" about the a%is is a ha$$ark of e#en functions.>ote a$so that a$$ the e%onents are e#en (the e%onent on the constant terbein" <eroE x = H1 = ).So /$$ $u" x in for x , an! si$ifyEf ( x ) = 3( x )2 + = 3( x 2) + = 3 x 2 + Iy fina$ e%ression is the sae thin" /! starte! &ith, &hich eans that f( x ) is e#en.

2-. "etermine a%geraica%%y whether &(C) 2C3 D !C is even, o, or neither.

4>SD0Ef "rah this, &i$$ see that it is Fsyetric about the ori"inFG that is, if start at a oint

on the "rah on one si!e of the y -a%is, an! !ra& a $ine fro that oint throu"h theori"in an! e%ten!in" the sae $en"th on the other si!e of the y -a%is, &i$$ "et toanother oint on the "rah.his syetry is a ha$$ark of o!! functions.>ote a$so that a$$ the e%onents are o!! (since the secon! ter is x = x 1). his is ausefu$ c$ue. shou$! e%ect this function to be o!!.f ( x ) = 2( x )3 ( x ) = 2( x 3) + x = 2 x 3 + x Iy fina$ e%ression is the e%act oosite of &hat starte! &ith, by &hich ean thatthe si"n on each ter has been chan"e! to its oosite, Just as if /! u$ti$ie!throu"h by 1E

http://www.purplemath.com/modules/symmetry.htm







f ( x ) = 1Bf ( x )C = B2 x 3 x C = 2 x 3 + x his eans that f ( x ) is o!!.

2+. "etermine a%geraica%%y whether &(C) 2C3 D 3C2 D !C 6 ! is even, o, or neither.

4>SD0E

his function is the su of the re#ious t&o functions.>ote that its "rah !oes not ha#e the syetry of either of the re#ious ones, nor are a$$ its e%onentseither e#en or o!!. &ou$! e%ect this function to be neither e#en nor o!!.F$$ $u" % in for %, an! si$ifyEf(–x) = 2(–x)3 – 3(–x)2 – 4(–x) + 4 = 2(–x3) – 3(x2) + 4x + 4 = –2x3 –3x2 + 4x + 4his is neither the sae thin" starte! &ith (nae$y, 2%3 3%2 % + ) nor the e%act oosite of &hat starte! &ith (nae$y, 2%3 + 3%2 + % ).his eans that f(%) is neither e#en nor o!!.

Remainder Theorem27. he e%ression %2 % + * $ea#es a reain!er of 2 &hen !i#i!e! by % 3. in! the#a$ue of .So$utionE @et f(%) = %2 % + *Ky the 0eain!er heore,f(3) = 2(3)2 3 + * = 2 = 15

2. in! the reain!er &hen %3 5% + 1 is !i#i!e! by % 2.So$utionE@et f(%) = %3 5% + 1hen f(%) is !i#i!e! by % 2, reain!er,0 = f(2) = (2)3 5(2) + 1 = 23

34. in! the reain!er &hen %3 5% + 1 is !i#i!e! by % + 3.

So$utionE@et f(%) = %3 5% + 1hen f(%) is !i#i!e! by % + 3, reain!er,0 = f(3) = (3)3 5(3) + 1 = 2

31. in! the reain!er &hen %3 5% + 1 is !i#i!e! by 2% 1.So$utionE@et f(%) = %3 5% + 1

2. 4 cosetics coany nee!s a stora"e bo% that has t&ice the #o$ue of its $ar"est bo%. ts $ar"estbo% easures 5 inches by inches by 3 inches. he $ar"er bo% nee!s to be a!e $ar"er by a!!in"the sae aount (an inte"er) to each to each !iension. in! the increase to each !iension.



hen f(%) is !i#i!e! by 2% 1, reain!er,

32. @et/s $ook at the o$ynoia$ (%) = %3 *% 6, an! $et/s !i#i!e by the $inear factor % So$utionE

() = ( )(()2 + () + ) + 3= ()(16 + 16 + ) + 3= + 3= 3

33. in! the reain!er if f(%) = %3 + 2%2 - 11% - 12 is !i#i!e! by % - 2.So$utionEhe reain!er theore te$$s us that h = 2, so e#a$uate f(2). e re$ace % &ith 2 in theo$ynoia$ as fo$$o&sEf(2) = 23 + 2(2)2 - 11(2) 12e then si$ifyE f(2) = 8 + 8 - 22 - 12 = -18. his te$$s us that if &e !i#i!e the o$ynoia$ by (% -2), &e &i$$ obtain a reain!er of -18.

3!. in! the reain!er 0 of the e:uation (%

L 5%3

+ %2

L 2% + 6) M (% + )So$utionE(% L 5%3 + %2 L 2% + 6) M (% + )

f(%) = % L 5%3 + %2 L 2% + 6f(L) = (-) L 5(L)3 + (L)2 L 2(L) + 6 = 66So the reain!er is 66.

3-. se the reain!er theore to fin! the reain!er of f(%) = 3%2 + 5% L 8 !i#i!e! by (% L 2).So$utionESince &e are !i#i!in" f(%) = 3%2 + 5% L 8 by (% L 2), &e $et %=2.'ence, the reain!er, 0 is "i#en byE0 = f(2) = 3(2)2 + 5(2) 8 = 1

3+. Ky usin" the reain!er theore, !eterine the reain!er &hen 3%3 L %2 L 2% + 5 is!i#i!e! by (% + ).So$utionESince &e are !i#i!in" by (%+), &e $et %=L. herefore the reain!er,0=f(L)=3(L)3L(L)2L2(L)+5=L12L16+8+5=L123

3. hat is the reain!er &hen 2%2-5%-1 !i#i!e! by %-3So$utionE

f(3) = 2(3)

2

-5(3)-1 = 18-15-1 = 2

37. se the reain!er theore to !eterine the reain!er &hen is !i#i!e! by E

A>'EF$:



3. se the reain!er theore to !eterine the reain!er &hen is !i#i!e! by E

A>'EF$:

!4. se the reain!er theore to !eterine the reain!er &hen is !i#i!e! by E

A>'EF$:

!1. se the reain!er theore to !eterine the reain!er &hen is !i#i!e! by E

A>'EF$:



!2. eterine the #a$ue of if !i#i!e! by "i#es a reain!er of 16.A>'EF$:

!3. ;a$cu$ate the #a$ue of if is !i#i!e! by an! "i#es areain!er of L2.A>'EF$:

!!. f is !i#i!e! by an! the reain!er is , fin! the #a$ue of .

!-. eterine the #a$ue of if is !i#i!e! by an! "i#es areain!er of 6.A>'EF$:



!+. f is !i#i!e! by an! the reain!er is , ca$cu$ate the #a$ue of .A>'EF$:

!. ;a$cu$ate the #a$ue of if is !i#i!e! by an! "i#es areain!er of 1.A>'EF$:

!7. f is !i#i!e! by an! the reain!er is , e%ress in ters of .A>'EF$:

!. hen the o$ynoia$ is !i#i!e! by or , the reain!er is 5.eterine the #a$ues of an! .A>'EF$:



Factor Theorem

-4. s (% + 1) a factor of f(%) = %3

+ 2%2

L 5% 6So$utionE n this case &e nee! to test the reain!er r=L1.0 = f(r) = f(L1)=(L1)3+2(L1)2L5(L1)L6=L1+2+5L6=herefore, since f(L1)=, &e conc$u!e that (%+1) is a factor of f(%).

-1. se the actor heore to !eci!e if (% L 2) is a factor of f(%) = % 5 L 2% + 3%3 L 6%2 L % + 8.So$utionEf(2) = (2)5 L 2(2) + 3(2)3 L 6(2)2 L (2) + 8 = Since f(2)=, &e can conc$u!e that (%L2) is a factor.

-2. sin" the factor theore, sho& that % + is a factor of f(%) = 5% + 16 %3 15%2 + 8% + 16So$utionEor %+ to be a factor, f(-) ust be e:ua$ to .;a$cu$ate f(-)f(%) = 5% + 16 %3 15%2 + 8% + 16f(-) = 5(-) + 16 (-)3 15(-)2 + 8(-) + 16f(-) = 5(256) + 16(-6) 15(16) + 8(-) + 16f(-) = 128 12 32 + 16 = Since f(-) = , % + is a factor of f(%). -3. se the factor theore to !eterine if % 1 is a factor of f(%) = 2% + 3%2 5% + *

So$utionEor % 1 to be a factor, f(1) ust be e:ua$ to .f(%) = 2% + 3%2 5% + *f(1) = 2(1) + 3(1)2 5(1) + *f(1) = 2 + 3 5 + *f(1) = *Since f(1) N , % 1 is not factor of f(%) = 2% + 3%2 5% + *

-!. eterine &hether % + 1 is a factor of 3% + %3 %2 + 3% + 2



So$utionE@et f(%) = 3% + %3 %2 + 3% + 2f(1) = 3(1) + (1)3 (1)2 +3(1) + 2f(1) = 3(1) + (1) 1 3 + 2 = herefore, % + 1 is a factor of f(%)

--. eterine &hether % + 1 is a factor of % 6 + 2%(% 1) So$utionE@et "(%) = %6 + 2%(% 1) "(1) = (1)6 + 2(1)( 2) = 1herefore, % + 1 is not a factor of "(%)

-+. se the factor theore to factorise %3 3%2 + co$ete$y.So$utionE@et a(%) = %3 3%2 + a(-1) = (-1) 3 3(-1) 2 + a(-1) = -1 3 + a(-1) =

herefore (% + 1) is one factor a(%) = (% + 1)(%2 % + )a(%) = (% + 1)(% -2)(% 2)a(%) = (% + 1)(% 2)2

-. eterine if % - 3 is a factor of %3 + 5%2 - 1*% - 21So$utionE3O 1 5 -1* -213 2 211 8 * Since the reain!er is , the !i#isor is a factorPSho& that % + 2 is a factor of 2%3 + 3%2 - 8% - 12 an! then factor co$ete$y.

So$utionE-2O 2 3 -8 -12- 2 122 -1 -6 (2%2 - % - 6)(% + 2) = (2% + 3)(% - 2)(% + 2)-7. Sho& that % + 3 an! % - are factors of % - 2%3 - 13%2 + 1% + 2 an! then factorco$ete$y.So$utionE-3O 1 -2 -13 1 2-3 15 -6 -21 -5 2 8

O 1 -5 2 8 - -81 -1 -2 (%2 - % - 2)(% + 3)(% - ) = (% -2)(% + 1)(% + 3)(% - )

Exponential Functions +4. You ecie to uy a use car that costs G14,444. YouHve hear that the car mayepreciate at a rate o& 14I per year. At that rate, what wi%% the car e worth in - yearsJ'o%ution:



+1. The initia% va%ue o& your car is G24,444. A&ter 1 year, the va%ue is G1-,444.

(a) Ehat is the percent ecreaseJ

() Kin the va%ue o& the car at this same rate a&ter - years &rom the initia% va%ue.'o%ution:



+2. Legan has G24,444 to invest &or - years an she &oun an interest rate o& -I. Mow much money wi%%she have at the en o& - years i& (a) the interest rate compouns month%y () the interest rate compouns semiannua%%yJ'o%ution:

+3. The initia% va%ue o& your car is G24,444. A&ter 1 year, the va%ue is G1-,444.

(a) Ehat is the percent ecreaseJ

() Kin the va%ue o& the car at this same rate a&ter - years &rom the initia% va%ue.'o%ution:

+!. Laison rea%%y wants to uy a car in ! years an she wants to start saving &or a ownpayment. 9& she eposits G3-44 now with interest compouning continuous%y at 3I, whatown payment wi%% she have &or her carJ'o%ution:



+-. A io%ogist is researching a new%yiscovere species o& acteria. At time t 4 hours, heputs one hunre acteria into what he has etermine to e a &avora%e growth meium. 'iChours %ater, he measures !-4 acteria. Assuming eCponentia% growth, what is the growthconstant Nk N &or the acteriaJ ($oun k to two ecima% p%aces.)'o%ution:or this e%ercise, the units on tie t &i$$ be hours, because the "ro&th is bein" easure! in ters ofhours. he be"innin" aount P is the aount at tie t = , so, for this rob$e, P = 1. he en!in"aount is A = 5 at t = 6. he on$y #ariab$e !on/t ha#e a #a$ue for is the "ro&th constant k , &hicha$so haens to be &hat / $ookin" for. So /$$ $u" in a$$ the kno&n #a$ues, an! then so$#e for the"ro&th constantE

A = Pekt 5 = 1e6k

.5 = e6k ln(.5) = 6k

ln(.5)Q = k = .256*56612...

advanced algeb project.docx

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