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Inleiding en herhaling

Module by: Free High School Science Texts Project. E-mail the author

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

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