Algebra 1 - Leaving Cert Higher Level Maths

1 Algebra 1 - Leaving cert Higher Level MathsAdam KellyNovember 27, ,andIhavemo di epro ofsprovidedforsomestatementsandtheorems, theydonotneedtob elearnedforthepurp ofsarejusttoincreasetheunderstandingofth estatementsb Polynomial :axxn+an 1xn 1+ +a2x2+a1x+a0 Whereai R;i= 0,1,2, .. , n;n ,withnon-zerocoe eci ctyp esofp olynomialswhichyouneedtoknow. Alinearp olynomialisofdegree1(ax+b) Aquadraticp olynomialisofdegree2(ax2+bx+c) Acubicp olynomialisofdegree3(ax3+bx2+cx+d)Therea realsonamesforp olynomialsbasedonhowmanytermstheyhave. Amonomialisap olynomialcontainingasingleterm. Abinomialisap olynomialcontainingtwoterms. Atrinomialisap (forexample4in4x3).definitionConstant Addition and Subtraction of Polynomial ExpressionsAddingandsubtractingp olynomialsinvolvesthecombinationoflikete rmsbyaddingtheirco e the following expression:(6x2 7x+ 4) + (7x2 9x+ 8)Solution(6x2 7x+ 4) + (7x2 9x+ 8)= (6x2+ 7x2) + ( 7x 9x) + (4 + 8)= 13x2 16x2+ 12example Multiplying Polynomial ExpressionsPolynomialsaremultipliedusing thedistributiveprop ertyofaddition:a(b+c) =ab+acSimplify the following expression.

2 (x+ 4)(2x+ 5)Solution(x+ 4)(2x+ 5)= 2x2+ 8x+ 5x+ 20= 2x2+ 13x+ 20example Dividing Polynomial ExpressionsThereareafewcasesforthedivisi onofp elow, er,thefollowingequalitiesdonotholdwhenth equotient' ,ifyouhavetheexpression(2x+ 2) (x+ 1),andyousimplifyitto2,youmustrememb erthatthisexpressionisnotequaltotwoifx= 1,asthatwouldb edivisionbyzeroandthusunde Denominator is a Factor of Each TermIfthedenominatorisafactorofeachtermo fthenumerator,thequotientcanb esimpli the following expression: 16x4+ 8x2+ 2x2xexample 16x4+ 8x2+ 2x2x= 16x42x+8x22x+2x2x= 7x3+ 4x+ Denominator is a Factor of the NumeratorIfthedenominatorisafactorofthen umerator,thequotientcanb esimpli the following expression:15x2+ 22x+ 85x+ 4 Solution15x2+ 22x+ 85x+ 4=(3x+ 2) (5x+ 4) 5x+ 4= 3x+ 2example Polynomial Long DivisionIftheothertwometho dsdon'twork,youcanusep the following expression:(x3+x2 2x) (x 1)Solutionx2+ 2xx 1)x3+x2 2x x3+x22x2 2x 2x2+ 2x0example x3+x2 2xx 1=x2+ 2xYoushouldnotethatitisp ossiblethatitdo esn'tdivideinevenly, the following expression.

3 X3+x2 1x 1 Solutionx2+ 2x+ 2x 1)x3+x2 1 x3+x22x2 2x2+ 2x2x 1 2x+ 21 x3+x2 1x 1=x2+ 2x+ 2 +1x 1example Synthetic DivisionIfthedenominatoroftheexpressioni sx a,wherea R, cessof ndingthero otsofap example using synthetic 2011201200 x3+x2 2xx 1=x2+ 2xexample Factorising Polynomial ortantpartoftheLCmathscourse, ortantthatyougetgo o datthisskillandb elow, Highest Common FactorThismetho dcanb edonebylo + 6x= 2x(x+ 3)15axy+ 2xyb+ 4xy= 2xy(7a+b+ 2)example GroupingAnexpressioncansometimesb the expression6x2y+ 3xy2 12x 6ySolution6x2y+ 3xy2 12x 6y= 3xy(2x+y) 6(2x+y)= (2x+y)(2xy 5)example Difference of Two SquaresDifference of Two b2canbefactoredinto(a+b)(a b).theorem (a+b)(a b) =a(a b) +b(a b)=a2 ab+ab b2=a2 b2 QED5x2 36 = (x+ 6)(x 6)4x2 81 = (2x+ 9)(2x 9)9x3 81x= 9x(x2 9) = 9x(x+ 3)(x 3)16x4 1 = (4x2+ 1)(4x2 1) = (4x2+ 1)(2x+ 1)(2x 1)example Difference of Two CubesDifference of Two b3canbefactoredinto(a b)(a2+ab+b2)theorem (a b)(a2+ab+b2) =a(a2+ab+b2) b(a2+ab+b2)= (a3+a2b+ab2) (a2b+ab2+b3)= (a3+ a2b+ab2) ( a2b+ab2+b3)=a3 b3 QEDx3 8 = (x 2)(x2+ 2x+ 4)125x3 27y3= (5x 3y)(25x2+ 15xy+ 9y2)example Sum of Two CubesSum Of Two +b3canbefactoredinto(a+b)(a2 ab+b2)theorem (a+b)(a2 ab+b2) =a(a2 ab+b2) +b(a2 ab+b2)=a3 a2b+ab2+ba2 ab2+b3=a3 (((((((((((a2b+ab2+ba2 ab2+b3=a3+b3 QED627x3+ 1 = (3x+ 1)(9x2 3x+ 1)81x3+ 192b3= (5x+ 6b)(25x2 30xb+ 36b2)))))))))))

4 Example Factoring With The Quadratic FormulaIfthepreviousmetho dsdon'tworkandyouwithtofactoraquadratic, youcanusethsfollowingmetho d(showninworkedexample).Factor3x2 17x+ 20 SolutionFirst,youmustlettheexpressionequ al0,andthensolveforx:3x2 17x+ 20 = 0= x=17 ( 17)2 4(3)(20)2(3)=17 76= x= 4orx=53 Fromthisyoucan ndthatifx= 4,thenx 3x= 5,then3x 5isafactor 3x2 17x+ 20 = (3x 5)(x 4)example Simplifying Algebraic FractionsAlgebraic :f(x)g(x)Wheref(x), g(x) edinmuchthesamewayasnumericalfractions,a ndtheyareadded,subtracted, cationoffractionscanb edescrib edwithanumb erofrules: Fractionscanb eaddedoncetheyhaveacommondenominator. Afractioncanb esimpli edi thenumeratoranddenominatorhaveacommonfac tor(thisissometimesreferredtoascancellin g).

5 7 Ifthedenominatorornumeratorcontainsafrac tion,theyshouldb ereducedb eforepro enotedthatinmostcasesitisnotnecessarytoe xpandfullytheexpressioninthedenominatoro fthefraction,andusuallycanb ertiescanb eusedduringthissimpli Of +cd=ad+bcbdtheorem +cd=ab dd+cd bb=adbd+bcbd=ad+bcbdQEDF ractions In The 1cd=ab (cd) 1=ab dc= the following expression:1x+ 2+1x+ 3 Solutionexample + 2+1x+ 3=(x+ 2) + (x+ 3)(x+ 2)(x+ 3)=2x+ 5(x+ 2)(x+ 3)Simplify the following expression:2x2 5x 34x2 1 Solution2x2 5x 34x2 1=(2x+ 1)(x 3)(2x+ 1)(2x 1)= (2x+ 1)(x 3) (2x+ 1)(2x 1)=x 32x 1example Binomial The Binomial CoefficientTounderstandthebinomialexpans iontheorem,youmustb efamiliarwiththebino-mialco e e e cient:(nr)Representsthenumb erofwaysyoucancho oserob jectsfromagroupofnob e cientiscalculatedas:(nr)=n!

6 R!(n r)!Thereisanumb erofprop (nr)=(nn r)theorem theorem has an intuitive proof (which you should read!), along with aformal proof. To understand how this works, imagine you havenballs, and you areasked to chooserof them to remove from the group. You could also think of thisas choosingn rof them to keep. Thus, choosingrobjects out ofnis equivalentto choosingn robjects out , you can show this theorem algebraically:(nr)=n!r!(n r)!= (nn r)=n!(n r)!(n (n r))!=n!(n r)!r!QEDI tisalsop ossibletocalculatethisco e cientinanalternateway,whichcanb eusefulforcalculatingthisco e cientwithoutacalculator.(nr)=n(n 1)(n 2)..(n (k 1))k(k 1)(k 2)..1 Without a calculator, find(74)Solution(74)=7 6 5 44 3 2 1=84024= 35example a calculator, find(1511)Solutionexample (1511)=(154)=15 14 13 124 3 2 1= The Binomial TheoremThe Binomial ,(a+b)n=n m=0(nm)an mbn=(n0)an+(n1)an 1b+(n2)an 2b2+(n3)an 3b3+ +(nn)bntheorem (a+b)n= (a+b)(a+b)(a+b).

7 (a+b) ntimes. Repeatedly using the distributiveproperty, we can see that for a termambn m, we must choosemof thenterms tocontribute anato the term, and then each of the othern mterms of the productsmust contribute abterm. Thus the coefficient of theambn mis the number of waysto choosemobjects from a set of sizen, which is(nm). For all possible values of0 m n, we can see that(a+b)n=n m=0(nm)ambn mQED5 Algebraic ,al lcoe e cientsonb othsides,thentheyareequalFind the values ofa, bandcgivenax2+bx+c= 3x2+ 2x+ 5 For all values 3, b= 2, c= 51(x+ 1)(x 1)=A(x+ 1)+B(x 1)For all values ofx, find values (x+ 1)(x 1)=A(x+ 1)+B(x 1)=A(x 1) +B(x+ 1)(x+ 1)(x 1)= 1 =A(x 1) +B(x+ 1)Sincethisequationistrueforallvaluesofx ,youcanpicktwovaluesforxandAandBcanb 1:1 =A(x 1) +B(x+ 1)=A(( 1) 1) +b(( 1) + 1)= 2A A= 1:1 =A(x 1) +B(x+ 1)=A(1 1) +b(() + 1)= 2B B=12,Givingthe nalanswer:A= 12, B=12example