Summer Assignment

MAT 405: Precalculus Essentials

1450 Newfield Avenue Stamford, CT 06905 (203) 322-3496

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Knowledge for the Journey

Summer Math Review Packet MAT 405: Pre-Calculus Essentials

Directions: There are 8 topics in this summer assignment. Take the assignment one topic at a time. For each topic,

1. 1) Read the notes. 2. 2) Answer the odds*** of the associated problems.

1. Show your work on the page below the problems and on the back if necessary. 2. Write your final answers on the answer sheet.

3. 3) Begin the next topic.

*** For Topics 4 and 6 you must do all of the problems.

This assignment is due on the first day of classes. It will count as your first Take Home Quiz and you will eventually have a test on these topics. Please use the internet or last year’s notes if you need extra help. Blank answers will not be accepted and will result in After School Study Hall.

To ic 1 Factorin

FACTORING POLYNOMIALS

Often we ,want to un-multþly or FACTOR a polynomi"l l(Ð This process involvgs

finding a constant and/orunótfto polynomial that evenly divides the given polynomial'

In formal mathematical terms, this means P(x) = q(*) r(x), where q and r are also

polynomials. For elementary algebra there are three general types of factoring'

l) Common term (finding thg largest common factor):

6x+ 18 = 6(x+ à) *tt.i. 6 is a coÍrmon factor ofboth terms'

2x3 -8x2 - 10x = 2x(x' - 4x-5) where 2x is the common factor'

2x, (x- 1)+ 7 (*-1) = (x -t)(Zx'+ 7) where x - I is the common factor.

2) Special products

a' -b'= (a+ U)(a - U)

x2 + 2xy + y' =(" * Y)'

x'-zxy*y'=(*-Y)'

3a) Trinomials in the form x2 + bx + c where the coefficient of x2 is 1'- / Consider xt +(d+e)x+d.e=(x+d)(x+e), where the coefficientof x is

the sum of two numbers d and e AND the constant is the protlugltof !h9same two numbers, d and e. A quick way to determine all of the possible

pairs of integers d and e is to fáctor the cdnstant in the original trinomial''Fo,

.*u-ple, tZis 1.12 , 2'6 , and 3'4 . The signs of the two numbers are

determinà by the combination you need to get the sum' The "$lln and

product,, upprourh to factoring trinomials is the same as solving a "Diamond

Þroblem" in CpVt't Algebra 1 course (see below)'

x2 + 8x+15 = (x+ 3)(x+ 5); 3+5 = 8, 3'5 = 15

x, -2x-15 = (x- S)(x+ 3);- 5 + 3= -2,- 5'3= -15

x, - 7 x * rz =(x - r) (x - q); -3 t (;4) = -2, (-:) ( -4) = tz

x' -25 = (x + S)(x - S)

9xz - 4y' =(:x + zy)þx-zv)x2+gx+16=(x+a)'xr-gx*16=(*-a)"

To ic 1 Factorin

Factor each pollmomial completely'

l. x2-x-42 2. 4x2-!8 3. 2x2+9x+9 4'

5. 6xz -x-15 6 4x2 -25 7.x2 -28x+I96 8'

g. x2 +18x+81 10' x2 +4x-21. 11' 3x2 +Zlx 12'

13. gxz -16 t4. 4xz +20x+25 15' x2 - 5x + 6 16'

17. 4xz +18 18. x2 -!2x+36 19. x'-3x-54 20'

21. Zxz +15x+ 18 22. L6x2 -l 23' x' - 14x+ 49 24'

25. 3x3-l¡xz-45x26. 3x2 +24 27. x2 +t6x+64

2x2 +3xy+Y2

7x2 -847

3x2 -2ox-325x3 + l5x2 -20x

6x2 -21

x2+8x+15

To ic 2 Solvin uadratic E uations bFactoring

ZERO PRODUCT PROPERTY AND QUADRATICSIf a' b = 0, then either a = 0 or b = 0.

Note that this property states that at least one of the factors MUST be zero. It isalso possible that all of the factors aÍe zero. This simple statement gives us apowerful result which is most often used with equations involving the products ofbinomials. For example, solve (x + 5)(x - 2) = 0 .

BytheZeroProductProperty, since (x+5Xx -2)=0, either x+5 =0 orx-2=0. Thus, x=-5 Or x=2.

TheZ.ero Product Property can be used to find where a quadratic crosses ttre x-axis.These points are the x-intercepts. kr the example above, they would be (-5, 0) and (2, 0)

To ic 2 Solvin uadratic E uationsFactorinq

Solve the following problems using theZero Product Property.

l. (x-2Xx+3)=g 2. 2x(x+5Xx+6)=0

3. (x*lSXx-3)=0 4. 4x2-5x-6=05. (2x-1Xx + 2)=0 '6. 2x(x-3)(x+ 4) = 0

7. 3x2-13x-10=0 8. 2x2-x=15

9. Y =x2-3x+212. Y =x2 -4x-515. Y=x2 -8x+16

10.y=x2-10x+2513. Y =N2 +2x-816.y=x2 -9

11. y =P -x- 12

14. y=x2 +6x+9

To ic3 uadratic Formula

Example 1

Solve x2 + 3x - 2 : 0 using the quadratic formula'

First, identiff the values for a, b, and c. In this case they are 1,3, atd-2, respectively'

Ñ;;¿ substitute these values into the quadratic formula'

-(3) t 32 4$)(1) -3t..n=

22(L)åXx

3= h7 -3- JtlThen split the numerator into the two values: xt 2- or x =

2

TI{E QUADRATIC FORMULAyou have used factoring and theZeroProduct Property to solve quadratic

equations. Vo,, ,* roií. g4y quadratic equation by using the QUADRATIC

F'ORMULA.

For example, suppose Z* + 7x - 6: 0' Here a : 3' b :7 ' aîd c: -6'

Substituting these values into the formula results in:

-b+ b2 - 4aclf xx?+bx*c=00 then x= 2a

+ q') -t *^[tnã x: 6x-7 *tI+ *:T

2(3)

Remernber that non-negative numbers have both a positive and negative square

root. The sign * repre-sents this fact for the square root in the formula and

allows us to write th'e equation once (representing two possible solutions) until

later in the solution Process'

Split the numerator into the two values: x : -+ or x: +

Thus the solution for the quadratic equation is: 7:l or -3'

To ic3 uadratic Formula

use the quadratic formula to solve each of the following equations.

1. *-x-6:0 2. x2+8x+15=0

3. *+t3x+42:0 4. x2- 10x+ 16=0

5. x2+5x+4:0 6. x2-9x+18=0

7. 5x2 -x-4:09. 6x2 _x_ 15 = 0

11. 3x2 + 5x-28:0

13. 4* -9xt 4:0

15. 20r'2 + 20x: I

17. 7v? + 28x: 0

19. 8x2 - 50:0

8. 4P - 1lx- 3 :0

10. 6x2+ l9x+ 15=0

12. 2* -x- 14:0

14. 2x2-5x*2=0

16. l3r.2 -76x:4

18. 5*:-125x

20. l5x2 :3

To ic 4 Com letin the S uarer Stap 1 Divide all terms by a (the coefficient of x2).

. step 2 Move the number term (c/a) to the right side of the equat¡on.

. Stcp 3 Complete the square on the left side of the equation and balance this by

adding the same value to the right side of the equation.

W€ now have something that looks l¡ke (x + p)2 = q, which can be solved rather eas¡¡yr

. gtep 4 Take the square root on both sides of the equat¡on.

. Step 5 Subtract the number that remains on the lefr side of the equation to ñnd

x.

Example 1: Solve x2 + 4x + I = 0

Sbp 1 can be sklpped ln thls examplc slnce thê coeficlent of x2 ls 1

Stcp 2 Move the number term to the rlght sfde of tâe equatlon:

ât, x2+4X=-1

Stcp 3 Complete üre sguare on the left slde of the equatlon and balance thls by addlng

the same number to the rlght side of the equatlon.

þn)z=(412)2=22=4

4t *2 +4x+ 4='L+4-Þ (* +2)2=3

StGp ¿l Take the squâre root on both sldes ofthe equat¡on:

â> x * 2 =*y'3 = *1.23 (to 2 àeiimals)

StGp 5 Subtract 2 from both sldes:

-f¡ x ='*.L73 - / ='3,73 or'CI,27

To ic 4 Com letin the S uare

Solve cach cqurüon by complcüng the ¡qurrc.

D b2 -6b -51 - -3 2) ^' -2în - 4! -7

3) k' +8&-36--5

5' h2 -L4k-?4 -4

4) 8r'+L6x-1fi)--2

6)3ø2-6a-79--7

Top ic 5 Laws of Exponents

Example 1

Simpliff: (z*y')(s*'yo)

Multiply the coefficients: 2'5 ' xyt ' *"yo = 10xy3 '*'yo

Add the exponents of x, then y: 10x1*2y3*a = 10.*'yt

Example 2Simpliff: #Divide the coeffici.rtr,

(toT)tf" = +l-

Subtract the expone nts: 2x2-syt2-7 - 2*-tyt oR +

Example 3

Simpliff: (r*"0)'

Cube each factor: ¡''(*')''(t-)' = zl (*')' (vo)'

Multiply the exponents: 27x6yr2

BASE, EXPONENT, AND VALUE

In the expression 25 ,2 is the base, 5 is the exponent, and the value is 32.

25 means 2. 2. 2.2.2 =32 x3 means x . x . x

LAWS OF EXPONENTS

Here are the basic patterns with examples:

1) xu'xb=xurb examples:x3'^4= x3+q-x7;

14

examples: xlo:x4=¡10-4=x6i þ =2-3

examples: (x+¡: - x43 - *tz; ç2x3)4 - 24xr2 - l6xtz

*-"=å and$=xb examPles: 3x-3y2 =!-.;

LAWS OF EXPONENTS

a7 ¡4 - ¡ll

xu

Fa-b2)

4)

=[

3) (xu)b

! = 27ty'v-

= xub

ToSimpliff each expression:

l. yt.y' 2. b4.b3.b2

ic 5 LawS of E onents

5. þu)o 6.m8

;3-

3. 'g6 .g2

7

4. ytt

8' (*'Y')'l2xe

7{

9('o )'

GT10.

14.

15x2y7

ïrfl n. (4c4Xac')(ra'c) t2. (r*'y')'

13. (+xy')(zy)' (+ 16. l4l'\.mt,/

20. þr")',(::r^)'8x'y'"

1)'

(zu7)(t"')--77-

17. (zu'*')"(z*o)' rr. (*)' 19. 6y"^8

12x3y1 )'

To ic 6 Evaluatin Functions

r Evaluata.P= 4r- 3 ¡tr= CI.

y=4{0}-J=0-3=-l

This ¡neans that lhe point (0, {} is on theline.

. Ev¡luetey={r-3 atx= 3.

y:4{3)-J=12-3:9

Then (3,9) is on lhe line.

r Evaluate"f[-3).

To evatuate a function, I do just what I did above: I plug in the given value furx. Her€. I am

supposed to evaluata at the value¡ = -3. The notalion ie different, butf-3)'mêan$ exac{y the

same thing ås "evatuate .flù = Jæ at* = -3"t

"rl4)= {z=,tñl= Jz:-g= ^lß =4

. Evaluets"fl3).

25- I25-9

=1ffi=4

To ic 6 Evaluatin FunctionsOt¡ún* þrn d, frtrrfu, Þ lrra @ ntul. Iæt g(r) o x2 - 5r + 2. Find úe 6llæ'ing:

r, gGll b.c(¿l

2. lÅlbl '2* +Z.Fi¡d o¡phof tho foltoniag:

r. f(-3) b. t(61

c. s{O

a J{-1}

d. dsl

d.t(q

3. I¡t g(x)'.f + ¡lx- 1, Find the follon'ing:

r. s(.{l b. øl8l

a. l"d llxl * 3t' - 5¡r. Find oa¡ù of thc following:

t f(21 b. /l-8)

a ct-l}

¿ fFl

d. c{4

d.t{-lt

To ic 7 Sim Radicalsa

RADICALSSometimes it is convenient to leave sqlrare roots in radical form instead of using

a calculator to find approximations (dãcimal values). Look for perfect squares

(i.e., 4, g, 16, ZS, X, ig,"') as factors of the number that is inside the radical

sign (radicand) andtake the squareroot of any perfeit s-quare factor' Multiply

the root of the p.rf..ffiare times the reduceá radical' When there is an existing

valuethatmultiptiestrleradical,multiplyanyroot(s)times.thatvalue.

For examPle:

J9 =3 5Jl=5'3=15

Jrg = Js2 =Js.Jl=3Jt 3\þ8 : zJ-+g'2-3'7Jl=2tJl

J8o =Jî6ó =Jt6..¡5 =+Js Jq5++Jn=JÇ ¿+qJa'5=3Ji+4'2J5=11J5

When there are no more perfect square factors inside the radical sign' the product

of the whole r"r"b;;i;;'fraction)ànd the remaining radical is said to be in

SIMPLE RADICAL FORM.

Simple radical form does not allow radicals in the denominator of a fraction' If

thereisandica]inthedenominator,RATIONALIZETHEDENOMINATORby multiplying the numerator and denominator of the fraction by the radical in

the original ¿.rro*i"ãtot. Then simpliff the remaining fraction' Examples:

2 z Jî zJt-6:=:-::-- --Jt "lz Jz 2

In the first examPle, Jl'Jl = Jq

J6 J6 = "$6 =6 and, l=1.

4Jl 4Jj J6 JTI 2J:n

-=:':-J6 Jo Jo 6õJ

-2 and f,=1. In the second examPle,

Example 1

Add 'ln + Jn - J ß. Factor each radical and simpliff'

Js 3 + "[q 3 - $6 3 - 3J3 + zJj - 4Jljr'Æ o' Jd

Example 2aJ 3J6 J6

JSimplify G Multþly by

J6Jø

and simpliff:ß

J6:Jo 62

RadicalsWrite each of the following radicals in simple radical form.

t. {24 2. Jß 3. \m

4. &r s. 1Æ 6. .Æ

7.'{ñ 8. J2ß q. .Æ'+JIs

n.Ja+.,Æ t+.Ju+Jgs ls.",fo-+Jn-Jn

ra.å'lzl

2r. 4Ji -#

To ic 7 Sima a

rc. J-ts-Js-Jn nr;

rs. I zov5

3

J¡.Æ

Jt

To ic 8 Sim Rational E ressronsa aa

SIMPLIFYING RATIONAL EXPRE SSIONS

RATIONAL EXPRESSIONS are fractions that have algebraic expressions in

their numerators and/or denominators. To simpliff rational expressions find

factors in the nurnerator and denominator that arethe same and then write them

as fractions equal to 1. For example,

å=t #=t ffi=l ffi=tNotice that the last two examples involved binomial sums and differences. Only

when sums or differences are exactly the same does the fraction equal 1' Rational

expressions such as the examples below CANNOT be simplifred:

(6+5)6

*3+yx3

xx+2

3x-22

Most problems that involve rational expressions will require that you factor the

numeiator and denominator. For example:

# = i Notice that I and ] each equal 1.

2.3.x2.x.v2 )= *çríp = # Notice that l, 5, and4: l.

ffi = *ltffi = 4 whereiå=t.

All three examples demonstrate that all parts of the numerator and denominator-

whether constants, monomials, binomials, or factorable trinomials:must be

written as products before you can look for factors that equal 1.

One special situation is shown in the following examples:,

I - -'l -I = -1 -*-? = -(**.') = -1-T- ' x x'+2 x'+z

5_x__(x-5)__1x-) x-)

Note that in all cases we assume the denominator does not equal zero.

t254

6x3yzl5xzya

Example 1

simplify ffi## completelY

Factor: wffiNote: ;, S , and fi= 1, so the

simplifiedformis 8, x#l ot -2.

Example 2

sirnpliff #. x2-6x+8

Factor: nax'+4K-12

Since #=t, # it the simplified

form, x* -6 or 2.

completely

(x-4Xx-2)(x+O(-;:Ð

To ic I Sim Rational E ressronsa

1. 2(x+3)4(x-2)

4. 4(x- 3Xx- 5)

6(x- 3)(x+2)

7. 24(y -4Xv-6)16(y+6X6-Y)

a

*-2)' 11'

2. 2(x-3)6(x+2)

5. 3(x -3Xa-x)15(x+ 3Xx- a)

8. 36(y + a)(y - 16)

6(x-2)(x+2)

6. 15(x - 1X7 - x)

25(x+ 1Xx- 7)

a

Simplifyeachofthefollowingrationalexp,ressioncompletely.Assumethedenominatorisnotequaltozøo.3. 2(x+3)(x-2)

10. (x+r)o(

32(y+16X16-Y) (x+3) (*-2

x+5 x-3 12. 2x,-5 (x+ 3

(x+5 (* - 3)' (zx- s) (x+3

9. (x+3 x-2

x-7 x+213.

(x+3)'(*-z)

(s-x)'(x -rf 14. (5-*)o( 3x - 1)2 15. rz

1* +5¡* 1* - z¡' (*- 5)n (zx-z)' 20(x-7 (x+2

16. 24( 3x-7)(x+1)6 17. * -t 18. x2-4(x+ 1)(x-2) (x+ 1)2(x-2)

zo(zx-t)

19. x2-4x2+x-6

20. x2-16x3 +9r? +2ox

'(x + t)'

Final Answers Topic 1 Factoring 1) 3) 5) 7) 9) 11) 13) 15) 17) 19) 21) 23) 25) 27)

Topic 2 Solving Quadratic Equations by Factoring 1) 11) 3) 13) 5) 15) 7) 9)

Topic 3 Quadratic Formula 1) 3) 5) 7) 9) 11) 13) 15) 17) 19)

Topic 4 Completing the Square 1) 2) 3) 4) 5) 6)

Topic 5 Laws of Exponents 1) 3) 5) 7) 9) 11) 13) 15) 17) 19)

Topic 6 Evaluating Functions 1a) 1b) 1c) 1d) 2a) 2b) 2c) 2d) 3a) 3b) 3c) 3d) 4a) 4b) 4c) 4d)

Topic 7 Simplifying Radicals 1) 3) 5) 7) 9) 11) 13) 15) 17) 19) 21)

Topic 8 Simplifying Rational Expressions 1) 19) 3) 5) 7) 9) 11) 13) 15) 17)