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Kyk weer na die oefening in deel B van die vorige aktiwiteit – het jy die probleme herken?

C Gemene faktore van veelterme

Presies dieselfde metode word gebruik as ons die gemene faktore van meer as twee terme moet vind.

  • Voorbeelde:

6x 3 – 3x 2 + 6x = 3x (2x 2 – x + 2)

ab 3 c – 3a 2 b 3 c + a 3 b 2 c = ab 2 c (b – 3ab + a 2 )

3a + 24a 2 + 6a 3 = 3a ( 1 + 8a + 2a 2 )

20x – 8x 2 + 16x 3 – 12x 4 +4x 5 = 4x (5 – 2x + 4x 2 – 3x 3 + x 4 )

As jy mooi kyk, sal jy oplet dat die terme wat in die hakies oorbly, nie meer enige gemene faktore het nie. Dis wat gebeur as die uitdrukking ten volle gefaktoriseer is. Jy moet altyd die grootste moontlike gemene faktor van al die terme uithaal.

Oefening:

Faktoriseer die volgende uitdrukkings volledig deur die grootste gemene faktor uit te haal:

  1. 12abc + 24ac
  2. 15xy – 21y
  3. 3abc + 18ab 2 c 3
  4. 8x 2 y 2 – 2x
  5. 2a 2 bc 2 + 4ab 2 c – 7abc
  6. 12a(bc) 2 – 8(abc) 3 + 4(ab) 2 c 3 – 20bc + 4a

Paaraktiwiteit:

Het jy opgelet dat in elke geval die aantal terme in die hakies na faktorisering presies dieselfde is as die aantal terme in die oorspronklike uitdrukking?

Verduidelik vir jou maat hoekom jy dink dat dit altyd so sal gebeur.

D Faktore van die verskil van kwadrate

In deel D van die vorige aktiwiteit moes jy hierdie drie pare tweeterme vermenigvuldig:

(a + b) (a – b) ,

(2y + 3) (2y – 3) en

(2a 2 + 3b) (2a 2 – 3b)

  • Hier is die oplossing:

(a + b) (a – b) = a 2 – b 2

(2y + 3) (2y – 3) = 4y 2 – 9

(2a 2 + 3b) (2a 2 – 3b) = 4a 4 – 9b 2

Let op dat die antwoorde ‘n baie spesifieke patroon aanneem: vierkant minus vierkant .

Ons noem dit die verskil van kwadrate of verskil van vierkante , en dit word so gefaktoriseer:

Eerste–vierkant minus tweede–vierkant

= ( eerste vierkant size 12{ sqrt { ital "eerste" - ital "vierkant"} } {} plus tweede vierkant size 12{ sqrt { ital "tweede" - ital "vierkant"} } {} ) ( eerste vierkant size 12{ sqrt { ital "eerste" - ital "vierkant"} } {} minus tweede vierkant size 12{ sqrt { ital "tweede" - ital "vierkant"} } {} )

  • Voorbeelde:

x 2 – 25 = (x + 5) (x – 5)

4 – b 2 = (2 + b) (2 – b)

9a 2 – 1 = (3a + 1) (3a – 1)

DIT WORD VAN JOU VERWAG OM GOED VERTROUD TE WEES MET DIE ALGEMEENSTE VIERKANTE EN HUL VIERKANTSWORTELS.

Hier is ‘n klompie belangrikes – voeg self ander by die lys.

2 2 = 4 3 2 = 9 (a 2 ) 2 = a4

(a 3 ) 2 = a 6

(½) 2 = ¼ 1 2 = 1

Oefening:

Faktoriseer volledig:

1. a 2 – b 2

  1. 4y 2 – 9
  2. 4a 4 – 9b 2
  3. 1 – x 2
  4. 25 – a 6
  5. a 8 – ¼
  6. 4a 2 b 2 – 81
  7. 0,25 – x 2 y 6

9. 2a 2 – 2b 2 (versigtig!)

E Gekombineerde gemene faktore en verskille van vierkante

Soos in die laaste probleem (9), is dit noodsaaklik om eers gemene faktore uit te haal, en om daarna die uitdrukking in die hakies te faktoriseer, indien moontlik.

  • Nog ‘n voorbeeld:

Faktoriseer 12ax 2 – 3ay 2

Herken eers die gemene faktor 3a, voor jy sê dat dit nie ‘n verskil van vierkante kan wees nie.

12ax 2 – 3ay 2 = 3a (4x 2 – y 2 ) Nou herken ons 4x 2 – y 2 as verskil van twee vierkante.

12ax 2 – 3ay 2 = 3a (4x 2 – y 2 ) = 3a(2x + y)(2x – y).

Oefening:

Faktoriseer volledig :

1. ax 2 – ay 4

2. a 3 – ab 2

3. 0,5a 2 x – 4,5b 2 x

4. a 5 b 3 c – abc

F Opeenvolgende verskille van vierkante

Hou jou oë oop en probeer hierdie tweeterm volledig faktoriseer: a 4 – b 4

Nou hierdie oefening – soos gewoonlik, faktoriseer volledig.

1. x 6 – 64

2. 1 – m 8

3. 3a 4 – 24b 8

4. x – x 9

G Faktore van drieterme

Bestudeer die antwoorde op hierdie vier probleme (uit ‘n vorige aktiwiteit). Die vereenvoudigde antwoorde het partykeer twee terme, partykeer drie terme en partykeer vier. Bespreek met ‘n maat wat hier aan die gang is en besluit wat die verskille veroorsaak.

Questions & Answers

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Source:  OpenStax, Wiskunde graad 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11055/1.1
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