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4.2 Faktorisering en breke  (Page 2/4)

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Daar is sekere wenke wat jy in gedagte kan hou:

  • As c positief is, moet albei die faktore van c positief of albei negatief wees. Die faktore is beide negatief indien b negatief is, en beide positief indien b positief is. As c negatief is, beteken dit slegs een van die faktore van c is negatief, en die ander een is positief.
  • Wanneer jy 'n antwoord gekry het, brei weer jou hakies uit net om te toets of dit reg uitwerk.

Vind die faktore van 3 x 2 + 2 x - 1 .

  1. Die kwadraat is in die regte vorm.

  2. ( x ) ( x )

    Skryf die stel faktore neer van a en c . Die moontlike faktore van a is: (1,3). The moontlike faktore van c is: (-1,1) of (1,-1).

    Skryf die groep opsies neer van die moontlike faktore van die kwadratiese uitdrukking a en c . Daar is twee moontlike oplossings.

    Opsie 1 Opsie 2
    ( x - 1 ) ( 3 x + 1 ) ( x + 1 ) ( 3 x - 1 )
    3 x 2 - 2 x - 1 3 x 2 + 2 x - 1
  3. ( x + 1 ) ( 3 x - 1 ) = x ( 3 x - 1 ) + 1 ( 3 x - 1 ) = ( x ) ( 3 x ) + ( x ) ( - 1 ) + ( 1 ) ( 3 x ) + ( 1 ) ( - 1 ) = 3 x 2 - x + 3 x - 1 = x 2 + 2 x - 1 .

  4. Die faktore van 3 x 2 + 2 x - 1 is ( x + 1 ) en ( 3 x - 1 ) .

Faktorisering van 'n kwadratiese drieterm

  1. Faktoriseer die volgende:
    (a) x 2 + 8 x + 15 (b) x 2 + 10 x + 24 (c) x 2 + 9 x + 8
    (d) x 2 + 9 x + 14 (e) x 2 + 15 x + 36 (f) x 2 + 12 x + 36
  2. Ontbind die volgende in faktore:
    1. x 2 - 2 x - 15
    2. x 2 + 2 x - 3
    3. x 2 + 2 x - 8
    4. x 2 + x - 20
    5. x 2 - x - 20

  3. Vind die faktore van die volgende drieterme:
    1. 2 x 2 + 11 x + 5
    2. 3 x 2 + 19 x + 6
    3. 6 x 2 + 7 x + 2
    4. 12 x 2 + 8 x + 1
    5. 8 x 2 + 6 x + 1

  4. Faktoriseer die volgende drieterme:
    1. 3 x 2 + 17 x - 6
    2. 7 x 2 - 6 x - 1
    3. 8 x 2 - 6 x + 1
    4. 2 x 2 - 5 x - 3

Faktorisering deur groepering

'n Verdere metode van faktorisering gebruik gemeenskaplike faktore. Ons weet dat die faktore van 3 x + 3 , 3 en ( x + 1 ) is. Soortelyk is die faktore van 2 x 2 + 2 x , 2 x en ( x + 1 ) . Gevolglik het ons 'n uitdrukking:

2 x 2 + 2 x + 3 x + 3

wat ons kan faktoriseer as:

2 x ( x + 1 ) + 3 ( x + 1 )

Nou kan ons sien daar is 'n ander gemene faktor: x + 1 . Gevolglik kan ons nou skryf:

( x + 1 ) ( 2 x + 3 )

Ons kry dit deur die x + 1 'uit te haal' (uit te deel) en te sien wat oorbly. Ons het + 2 x uit die eerste term en + 3 uit die tweede term. Dit word genoem faktorisering deur groepering .

Vind die faktore van 7 x + 14 y + b x + 2 b y deur groepering

  1. Daar is geen algemene gemeenskaplike faktore nie.

  2. 7 is 'n gemene faktor van die eerste twee terme en b is 'n gemene faktor van die tweede twee terme.

  3. 7 x + 14 y + b x + 2 b y = 7 ( x + 2 y ) + b ( x + 2 y )

  4. x + 2 y is 'n gemeenskaplike faktor.

  5. 7 ( x + 2 y ) + b ( x + 2 y ) = ( x + 2 y ) ( 7 + b )

  6. Die faktore van 7 x + 14 y + b x + 2 b y is ( 7 + b ) en ( x + 2 y )

Khan akademie video oor faktorisering van 'n drieterm deur groepering

Faktorisering deur groepering

  1. Faktoriseer deur groepering: 6 x + a + 2 a x + 3
  2. Faktoriseer deur groepering: x 2 - 6 x + 5 x - 30
  3. Faktoriseer deur groepering: 5 x + 10 y - a x - 2 a y
  4. Faktoriseer deur groepering: a 2 - 2 a - a x + 2 x
  5. Faktoriseer deur groepering: 5 x y - 3 y + 10 x - 6

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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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