FACTORING TECH

• Diff of squares = factor by gcf

  Coeff numerical val is 1= usual factor processes

  coeff num val is not 1 = ac method

  1. Mult num coeff a and c
  2. get values that when added is b
  3. factor by gcf

  if cannot perform ac due to c being large amount = substitution

  set A to middle term degree

  set A2 for a term degree

  -2x6 + 6x3

  A= x3

  A2= x3)(x3) = x6

  get gcf

  plot the As

  factor the poly with th As

  after factored bring back true val of A
• by grouping:

  separate poly two eq to 2 bi

  numericak coeff/constant both eq

  if same ratio then factor by grouping via separate gcf taking