Case 1. Factoring trinomial type #f(x) = x^2 + bx + c#.
The factored trinomial will have the form: #f(x) = (x + p)(x + q)#.
The new AC Method finds #2# numbers #p and q# that satisfy these 3 conditions:
- The product #p*q = a*c#. (When #a = 1#, this product is #c#)
- The sum #(p + q) = b#
- Application of the rule of Signs for real roots.
Reminder of the Rule of Signs.
- When #a and c# have different signs, #p and q# have opposite signs.
- When #a and c# have the same sign, #p and q# have the same sign.
New AC Method.
To find #p and q#, compose factor pairs of #c#, and in the same time, apply the Rule of Signs . The pair whose sum equals to #(-b)#, or #(b)#, gives #p and q#.
Example 1. Factor #f(x) = x^2 + 31x + 108. #
Solution. #p and q# have the same sign. Compose factor pairs of #c = 108#. Proceed: #...(2, 54), (3, 36), (4, 27)#. The last sum is #4 + 27 = 31 = b#. Then, #p = 4 and q = 27#.
Factoring form: #f(x) = (x + 4)(x + 27)#
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CASE 2 . Factor trinomial standard type #f(x) = ax^2 + bx + c# (1)
Bring back to Case 1.
Convert #f(x)# to #f'(x) = x^2 + bx + a*c = (x + p')(x + q')#. Find #p' and q'# by the method mentioned in Case 1 .
Then divide #p' and q'# by #(a)# to get #p and q# for trinomial (1).
Example 2 . Factor #f(x) = 8x^2 + 22x - 13 = 8(x + p)(x + q)# (1).
Converted trinomial:
#f'(x) = x^2 + 22x - 104 = (x + p')(x + q')# (2).
#p' and q'# have opposite signs. Compose factor pairs of #(ac = -104) --> ... (-2, 52), (-4, 26)#. This last sum is #(26 - 4 = 22 = b)#. Then, #p' = -4 and q' = 26#.
Back to the original trinomial (1):
#p = (p')/a = -4/8 = -1/2 and q = (q')/a = 26/8 = 13/4#.
Factoring form
#f(x) = 8(x - 1/2)(x + 13/4) = (2x - 1)(4x + 13).#
This new AC Method avoids the lengthy factoring by grouping.
