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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
Naam Voorbeelde (geskei deur kommas)
term a · x k , b · x , c m , d · y p , e · y , f
uitdrukking a · x k + b · x + c m , d · y p + e · y + f
koëffisiënte a , b , d , e
eksponent (of indeks) k , p
grondtal x , y , c
konstante a , b , c , d , e , f
veranderlike x , y
vergelyking a · x k + b · x + c m = 0
ongelykheid 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 ( a x + b ) en ( c x + 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

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

  1. ( 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
    .

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

Gestel die twee terme is a x + b en a x - b , dan is hulle produk:

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

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 ) geskryf as 2 x + 2 . Faktorisering sal dus begin met 2 x + 2 en eindig met 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, 2 x - 6 x 2 kan as volg gefaktoriseer word:

2 x - 6 x 2 = 2 x ( 1 - 3 x )

Ondersoek: gemeenskaplike faktore

Vind die grootste gemene faktore van die volgende pare terme:

(a) 6 y ; 18 x (b) 12 m n ; 8 n (c) 3 s t ; 4 s u (d) 18 k l ; 9 k p (e) a b c ; a c
(f) 2 x y ; 4 x y z (g) 3 u v ; 6 u (h) 9 x y ; 15 x z (i) 24 x y z ; 16 y z (j) 3 m ; 45 n

Verskil van twee kwadrate

Ons het gesien dat:

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

In [link] 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

gefaktoriseer kan word as:

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

Dus,

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

Byvoorbeeld, x 2 - 16 kan geskryf word as ( x 2 - 4 2 ) wat die verskil is tussen twee kwadrate. Dus, die faktore van x 2 - 16 is ( x - 4 ) en ( x + 4 ) .

Faktoriseer volledig: b 2 y 5 - 3 a b y 3

  1. b 2 y 5 - 3 a b y 3 = b y 3 ( b y 2 - 3 a )

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


  1. ( a - 4 ) is die gemene faktor
    3 a ( a - 4 ) - 7 ( a - 4 ) = ( a - 4 ) ( 3 a - 7 )

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

  1. 5 ( a - 2 ) - b ( 2 - a ) = 5 ( a - 2 ) - [ - b ( a - 2 ) ] = 5 ( a - 2 ) + b ( a - 2 ) = ( a - 2 ) ( 5 + b )

Hersien

  1. Vind die produkte / Verwyder die hakies:
    (a) 2 y ( y + 4 ) (b) ( y + 5 ) ( y + 2 ) (c) ( y + 2 ) ( 2 y + 1 )
    (d) ( y + 8 ) ( y + 4 ) (e) ( 2 y + 9 ) ( 3 y + 1 ) (f) ( 3 y - 2 ) ( y + 6 )


  2. Faktoriseer:
    1. 2 l + 2 w
    2. 12 x + 32 y
    3. 6 x 2 + 2 x + 10 x 3
    4. 2 x y 2 + x y 2 z + 3 x y
    5. - 2 a b 2 - 4 a 2 b


  3. Faktoriseer volledig:
    (a) 7 a + 4 (b) 20 a - 10 (c) 18 a b - 3 b c
    (d) 12 k j + 18 k q (e) 16 k 2 - 4 k (f) 3 a 2 + 6 a - 18
    (g) - 6 a - 24 (h) - 2 a b - 8 a (i) 24 k j - 16 k 2 j
    (j) - a 2 b - b 2 a (k) 12 k 2 j + 24 k 2 j 2 (l) 72 b 2 q - 18 b 3 q 2
    (m) 4 ( y - 3 ) + k ( 3 - y ) (n) a ( a - 1 ) - 5 ( a - 1 ) (o) b m ( b + 4 ) - 6 m ( b + 4 )
    (p) a 2 ( a + 7 ) + a ( a + 7 ) (q) 3 b ( b - 4 ) - 7 ( 4 - b ) (r) a 2 b 2 c 2 - 1


Questions & Answers

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hydrothermal synthesis
ISHFAQ
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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 10) [caps]. OpenStax CNX. Aug 04, 2011 Download for free at http://cnx.org/content/col11328/1.4
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