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Inleiding

In hierdie hoofstuk sal jy leer hoe om met algebraïese uitdrukkings te werk. Hersiening van vorige faktorisering en vermenigvuldiging van uitdrukkings sal dus nodig wees voordat die nuwe leerstof uitgebrei word vir Graad 10.

Hersiening van vorige Werk

Die volgende behoort bekend te wees, maar ons gee 'n paar voorbeelde ter herinnering.

Dele van Uitdrukkings

Wiskundige uitdrukkings is soos sinne en elke deel het 'n spesifieke naam. Jy behoort vertroud te wees met die volgende name wat die dele van wiskundige uitdrukkings beskryf.

a · x k + b · x + c m = 0 d · y p + e · y + f 0 a · x k + b · x + c m = 0 d · y p + e · y + f 0
(1)
Table 1
Naam Voorbeelde (geskei deur kommas)
term a·xka·xk ,b·xb·x, cmcm, d·ypd·yp, e·ye·y, ff
                    
uitdrukking a·xk+b·x+cma·xk+b·x+cm, d·yp+e·y+fd·yp+e·y+f
koëffisiënte aa, bb, dd, ee
eksponent (of indeks) kk, pp
grondtal xx, yy, cc
konstante aa, bb, cc, dd, ee, ff
veranderlike xx, yy
vergelyking a · x k + b · x + c m = 0 a · x k + b · x + c m = 0
ongelykheid d · y p + e · y + f 0 d · y p + e · y + f 0
binomiaal uitdrukking met twee terme
trinomiaal uitdrukking met drie terme

Produk van twee Binomiale

'n Binomiaal is 'n wiskundige uitdrukking met twee terme, soos (ax+b)(ax+b) en (cx+d)(cx+d). As hierdie twee binomiale vermenigvuldig word, is die volgende die resultaat:

( a · x + b ) ( c · x + d ) = ( a x ) ( c · x + d ) + b ( c · x + d ) = ( a x ) ( c x ) + ( a x ) d + b ( c x ) + b · d = a x 2 + x ( a d + b c ) + b d ( a · x + b ) ( c · x + d ) = ( a x ) ( c · x + d ) + b ( c · x + d ) = ( a x ) ( c x ) + ( a x ) d + b ( c x ) + b · d = a x 2 + x ( a d + b c ) + b d
(2)

Exercise 1: Produk van twee binomiale

Vind die produk van (3x-2)(5x+8)(3x-2)(5x+8).

Solution
  1. Step 1. Vermenigvuldig uit en vereenvoudig :
    ( 3 x - 2 ) ( 5 x + 8 ) = ( 3 x ) ( 5 x ) + ( 3 x ) ( 8 ) + ( - 2 ) ( 5 x ) + ( - 2 ) ( 8 ) = 15 x 2 + 24 x - 10 x - 16 = 15 x 2 + 14 x - 16 ( 3 x - 2 ) ( 5 x + 8 ) = ( 3 x ) ( 5 x ) + ( 3 x ) ( 8 ) + ( - 2 ) ( 5 x ) + ( - 2 ) ( 8 ) = 15 x 2 + 24 x - 10 x - 16 = 15 x 2 + 14 x - 16
    (3)
    .

Die produk van twee identiese binomiale, is bekend as die kwadraat (of vierkant) van binomiale en word geskryf as:

( a x + b ) 2 = a 2 x 2 + 2 a b x + b 2 ( a x + b ) 2 = a 2 x 2 + 2 a b x + b 2
(4)

Gestel die twee terme is ax+bax+b

 
en ax-bax-b
 
, dan is hulle produk:

( a x + b ) ( a x - b ) = a 2 x 2 - b 2 ( a x + b ) ( a x - b ) = a 2 x 2 - b 2
(5)

Dit staan bekend as die verskil van twee kwadrate (of vierkante).

Faktorisering

Faktorisering is die omgekeerde proses van die uitbreiding van hakies. Byvoorbeeld, as hakies uitgebrei word, word 2(x+1)2(x+1) geskryf as 2x+22x+2. Faktorisering sal dus begin met 2x+22x+2

 
en eindig met 2(x+1)2(x+1). In vorige grade het ons gefaktoriseer deur die uithaal van gemeenskaplike faktore en die verskil tussen twee vierkante.

Gemeenskaplike Faktore

Faktorisering deur die uithaal van gemeenskaplike faktore, is gebaseer daarop dat daar faktore is wat in al die terme voorkom. Byvoorbeeld, 2x-6x22x-6x2

 
kan as volg gefaktoriseer word:

2 x - 6 x 2 = 2 x ( 1 - 3 x ) 2 x - 6 x 2 = 2 x ( 1 - 3 x )
(6)
Ondersoek: Gemeenskaplike Faktore

Vind die grootste gemene faktore van die volgende pare terme:

Table 2
(a) 6y;18x6y;18x (b) 12mn;8n12mn;8n (c) 3st;4su3st;4su (d) 18kl;9kp18kl;9kp (e) abc;acabc;ac
(f) 2xy;4xyz2xy;4xyz
     
(g) 3uv;6u3uv;6u (h) 9xy;15xz9xy;15xz
     
(i) 24xyz;16yz24xyz;16yz
     
(j) 3m;45n3m;45n

Verskil van twee Kwadrate

Ons het gesien dat:

( a x + b ) ( a x - b ) = a 2 x 2 - b 2 ( a x + b ) ( a x - b ) = a 2 x 2 - b 2
(7)

In Equation 7 dui die = teken aan dat die twee kante altyd gelyk sal wees. Dit beteken dat 'n uitdrukking in die vorm:

a 2 x 2 - b 2 a 2 x 2 - b 2
(8)

gefaktoriseer kan word as:

( a x + b ) ( a x - b ) ( a x + b ) ( a x - b )
(9)

Dus,

a 2 x 2 - b 2 = ( a x + b ) ( a x - b ) a 2 x 2 - b 2 = ( a x + b ) ( a x - b )
(10)

Byvoorbeeld, x2-16x2-16

 
kan geskryf word as (x2-42)(x2-42) wat die verskil is tussen twee kwadrate. Dus, die faktore van x2-16x2-16
 
is (x-4)(x-4) en (x+4)(x+4).

Exercise 2: Faktorisering

Faktoriseer volledig: b2y5-3aby3b2y5-3aby3

Solution
  1. Step 1. Vind die gemene faktore: :
    b 2 y 5 - 3 a b y 3 = b y 3 ( b y 2 - 3 a ) b 2 y 5 - 3 a b y 3 = b y 3 ( b y 2 - 3 a )
    (11)
Exercise 3: Faktoriseer binomiale met gemeenskaplike hakies:

Faktoriseer volledig: 3a(a-4)-7(a-4)3a(a-4)-7(a-4)

Solution
  1. Step 1. Vind die gemene faktore :
    (a-4)(a-4) is die gemene faktor
    3 a ( a - 4 ) - 7 ( a - 4 ) = ( a - 4 ) ( 3 a - 7 ) 3 a ( a - 4 ) - 7 ( a - 4 ) = ( a - 4 ) ( 3 a - 7 )
    (12)
Exercise 4: Faktorisering wat die omruiling van getalle in hakies benodig:

Faktoriseer 5(a-2)-b(2-a)5(a-2)-b(2-a)

Solution
  1. Step 1. Let daarop dat (2-a)=-(a-2)(2-a)=-(a-2) :
    5 ( a - 2 ) - b ( 2 - a ) = 5 ( a - 2 ) - [ - b ( a - 2 ) ] = 5 ( a - 2 ) + b ( a - 2 ) = ( a - 2 ) ( 5 + b ) 5 ( a - 2 ) - b ( 2 - a ) = 5 ( a - 2 ) - [ - b ( a - 2 ) ] = 5 ( a - 2 ) + b ( a - 2 ) = ( a - 2 ) ( 5 + b )
    (13)
Hersien
  1. Vind die produkte / Verwyder die hakies:
    Table 3
    (a) 2y(y+4)2y(y+4)(b) (y+5)(y+2)(y+5)(y+2)(c) (y+2)(2y+1)(y+2)(2y+1)
    (d) (y+8)(y+4)(y+8)(y+4)(e) (2y+9)(3y+1)(2y+9)(3y+1)(f) (3y-2)(y+6)(3y-2)(y+6)


    Kliek hier vir die oplossing
  2. Faktoriseer:
    1. 2l+2w2l+2w
    2. 12x+32y12x+32y
    3. 6x2+2x+10x36x2+2x+10x3
    4. 2xy2+xy2z+3xy2xy2+xy2z+3xy
    5. -2ab2-4a2b-2ab2-4a2b


    Klier hier vir die oplossing
  3. Faktoriseer volledig:
    Table 4
    (a) 7a+47a+4(b) 20a-1020a-10(c) 18ab-3bc18ab-3bc
    (d) 12kj+18kq12kj+18kq(e) 16k2-4k16k2-4k(f) 3a2+6a-183a2+6a-18
    (g) -6a-24-6a-24(h) -2ab-8a-2ab-8a(i) 24kj-16k2j24kj-16k2j
    (j) -a2b-b2a-a2b-b2a(k) 12k2j+24k2j212k2j+24k2j2(l) 72b2q-18b3q272b2q-18b3q2
    (m) 4(y-3)+k(3-y)4(y-3)+k(3-y)(n) a(a-1)-5(a-1)a(a-1)-5(a-1)(o) bm(b+4)-6m(b+4)bm(b+4)-6m(b+4)
    (p) a2(a+7)+a(a+7)a2(a+7)+a(a+7)(q) 3b(b-4)-7(4-b)3b(b-4)-7(4-b)(r) a2b2c2-1a2b2c2-1


    Kliek hier vir die oplossing

Meer Produkte

Figure 1
Khan Akademie video oor die produk van polinoomuitdrukkings

Ons het gesien hoe om twee binomiale te vermenigvuldig in die afdeling "Produk van twee Binomiale". In hierdie gedeelte, gaan ons leer hoe om 'n binomiaal (uitdrukking met twee terme) met 'n trinomiaal of drieterm (uitdrukking met drie terme) te vermenigvuldig. Gelukkig gebruik ons dieselfde metode as om twee binomiaaluitdrukkings te vermenigvuldiging.

Byvoorbeeld, vermenigvuldig 2x+12x+1 met x2+2x+1x2+2x+1

( 2 x + 1 ) ( x 2 + 2 x + 1 ) = 2 x ( x 2 + 2 x + 1 ) + 1 ( x 2 + 2 x + 1 ) ( pas distributiewe eienskap toe ) = [ 2 x ( x 2 ) + 2 x ( 2 x ) + 2 x ( 1 ) ] + [ 1 ( x 2 ) + 1 ( 2 x ) + 1 ( 1 ) ] = 2 x 3 + 4 x 2 + 2 x + x 2 + 2 x + 1 ( brei die hakies uit ) = 2 x 3 + ( 4 x 2 + x 2 ) + ( 2 x + 2 x ) + 1 ( groepeer soortgelyke terme om te vereenvoudig ) = 2 x 3 + 5 x 2 + 4 x + 1 ( vereenvoudig om 'n finale antwoord te gee ) ( 2 x + 1 ) ( x 2 + 2 x + 1 ) = 2 x ( x 2 + 2 x + 1 ) + 1 ( x 2 + 2 x + 1 ) ( pas distributiewe eienskap toe ) = [ 2 x ( x 2 ) + 2 x ( 2 x ) + 2 x ( 1 ) ] + [ 1 ( x 2 ) + 1 ( 2 x ) + 1 ( 1 ) ] = 2 x 3 + 4 x 2 + 2 x + x 2 + 2 x + 1 ( brei die hakies uit ) = 2 x 3 + ( 4 x 2 + x 2 ) + ( 2 x + 2 x ) + 1 ( groepeer soortgelyke terme om te vereenvoudig ) = 2 x 3 + 5 x 2 + 4 x + 1 ( vereenvoudig om 'n finale antwoord te gee )
(14)

Tip: Vermenigvuldiging van Binomiaal met Trinomiaal:

In die vermenigvuldiging van die binomiaal A+BA+B met die trinomiaal C+D+EC+D+E, is die heel eerste stap om die distributiewe wet toe te pas:

( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E ) ( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E )
(15)

As jy dit onthou, sal jy nie 'n fout maak nie!

Exercise 5: Vermenigvuldiging van Binomiaal met Trinomiaal

Vermenigvuldig x-1x-1 met x2-2x+1x2-2x+1

Solution

  1. Step 1. Bepaal wat is gegee en wat word gevra :

    Twee uitdrukkings word gegee: 'n binomiaal, x-1x-1, en 'n trinomiaal, x2-2x+1x2-2x+1. Ons moet hulle met mekaar vermenigvuldig.

  2. Step 2. Bepaal hoe om die probleem aan te pak. :

    Pas die distributiewe wet toe en vereenvoudig daarna.

  3. Step 3. Los die probleem op :
    ( x - 1 ) ( x 2 - 2 x + 1 ) = x ( x 2 - 2 x + 1 ) - 1 ( x 2 - 2 x + 1 ) ( pas die distributiewe wet toe ) = [ x ( x 2 ) + x ( - 2 x ) + x ( 1 ) ] + [ - 1 ( x 2 ) - 1 ( - 2 x ) - 1 ( 1 ) ] = x 3 - 2 x 2 + x - x 2 + 2 x - 1 ( brei die hakies uit ) = x 3 + ( - 2 x 2 - x 2 ) + ( x + 2 x ) - 1 ( groepeer soorgelyke terme saam om te vereenvoudig ) = x 3 - 3 x 2 + 3 x - 1 ( vereenvoudig om die finale antwoord te kry ) ( x - 1 ) ( x 2 - 2 x + 1 ) = x ( x 2 - 2 x + 1 ) - 1 ( x 2 - 2 x + 1 ) ( pas die distributiewe wet toe ) = [ x ( x 2 ) + x ( - 2 x ) + x ( 1 ) ] + [ - 1 ( x 2 ) - 1 ( - 2 x ) - 1 ( 1 ) ] = x 3 - 2 x 2 + x - x 2 + 2 x - 1 ( brei die hakies uit ) = x 3 + ( - 2 x 2 - x 2 ) + ( x + 2 x ) - 1 ( groepeer soorgelyke terme saam om te vereenvoudig ) = x 3 - 3 x 2 + 3 x - 1 ( vereenvoudig om die finale antwoord te kry )
    (16)
  4. Step 4. Skryf die finale antwoord neer :

    Die produk van x-1x-1 en x2-2x+1x2-2x+1 is x3-3x2+3x-1x3-3x2+3x-1.

Exercise 6: Som van Derdemagte

Vind die produk van x+yx+y

 
en x2-xy+y2x2-xy+y2.

Solution

  1. Step 1. Bepaal wat is gegee en wat word gevra :

    Twee uitdrukkings word gegee: 'n binomiaal, x+yx+y, en 'n trinomiaal, x2-xy+y2x2-xy+y2.

     
    Ons moet hulle met mekaar vermenigvuldig.

  2. Step 2. Bepaal hoe om die probleem aan te pak :

    Pas die distributiewe wet toe en vereenvoudig dan verder.

  3. Step 3. Los die probleem op :
    ( x + y ) ( x 2 - x y + y 2 ) = x ( x 2 - x y + y 2 ) + y ( x 2 - x y + y 2 ) ( pas die distributiewe wet toe ) = [ x ( x 2 ) + x ( - x y ) + x ( y 2 ) ] + [ y ( x 2 ) + y ( - x y ) + y ( y 2 ) ] = x 3 - x 2 y + x y 2 + y x 2 - x y 2 + y 3 ( brei die hakies uit ) = x 3 + ( - x 2 y + y x 2 ) + ( x y 2 - x y 2 ) + y 3 ( groepeer soortgelyke terme om te vereenvoudig ) = x 3 + y 3 ( vereenvoudig om die finale antwoord te kry ) ( x + y ) ( x 2 - x y + y 2 ) = x ( x 2 - x y + y 2 ) + y ( x 2 - x y + y 2 ) ( pas die distributiewe wet toe ) = [ x ( x 2 ) + x ( - x y ) + x ( y 2 ) ] + [ y ( x 2 ) + y ( - x y ) + y ( y 2 ) ] = x 3 - x 2 y + x y 2 + y x 2 - x y 2 + y 3 ( brei die hakies uit ) = x 3 + ( - x 2 y + y x 2 ) + ( x y 2 - x y 2 ) + y 3 ( groepeer soortgelyke terme om te vereenvoudig ) = x 3 + y 3 ( vereenvoudig om die finale antwoord te kry )
    (17)
  4. Step 4. Skryf die finale antwoord neer :

    Die produk van x+yx+y

     
    en x2-xy+y2x2-xy+y2
     
    is x3+y3x3+y3.

Tip:

Ons het gesien dat:
( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3
(18)

Dit staan bekend as die som van derdemagte.

Ondersoek: Verskil van Derdemagte

Toon aan dat die verskil van derdemagte (x3-y3x3-y3

 
) gegee word deur die produk van x-yx-y
 
en x2+xy+y2x2+xy+y2.

Produkte

  1. Vind die produk van:
    Table 5
    (a) (-2y2-4y+11)(5y-12)(-2y2-4y+11)(5y-12)(b) (-11y+3)(-10y2-7y-9)(-11y+3)(-10y2-7y-9)
    (c) (4y2+12y+10)(-9y2+8y+2)(4y2+12y+10)(-9y2+8y+2)(d) (7y2-6y-8)(-2y+2)(7y2-6y-8)(-2y+2)
    (e) (10y5+3)(-2y2-11y+2)(10y5+3)(-2y2-11y+2)(f) (-12y-3)(12y2-11y+3)(-12y-3)(12y2-11y+3)
    (g) (-10)(2y2+8y+3)(-10)(2y2+8y+3)(h) (2y6+3y5)(-5y-12)(2y6+3y5)(-5y-12)
    (i) (6y7-8y2+7)(-4y-3)(-6y2-7y-11)(6y7-8y2+7)(-4y-3)(-6y2-7y-11)(j) (-9y2+11y+2)(8y2+6y-7)(-9y2+11y+2)(8y2+6y-7)
    (k) (8y5+3y4+2y3)(5y+10)(12y2+6y+6)(8y5+3y4+2y3)(5y+10)(12y2+6y+6)(l) (-7y+11)(-12y+3)(-7y+11)(-12y+3)
    (m) (4y3+5y2-12y)(-12y-2)(7y2-9y+12)(4y3+5y2-12y)(-12y-2)(7y2-9y+12)(n) (7y+3)(7y2+3y+10)(7y+3)(7y2+3y+10)
    (o) (9)(8y2-2y+3)(9)(8y2-2y+3)(p) (-12y+12)(4y2-11y+11)(-12y+12)(4y2-11y+11)
    (q) (-6y4+11y2+3y)(10y+4)(4y-4)(-6y4+11y2+3y)(10y+4)(4y-4)(r) (-3y6-6y3)(11y-6)(10y-10)(-3y6-6y3)(11y-6)(10y-10)
    (s) (-11y5+11y4+11)(9y3-7y2-4y+6)(-11y5+11y4+11)(9y3-7y2-4y+6)(t) (-3y+8)(-4y3+8y2-2y+12)(-3y+8)(-4y3+8y2-2y+12)


    Kliek hier vir die oplossing

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