Factoring Trinomials

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Factoring trinomials - Factors of trinomial expressions when the coefficient of the highest power is not unity.

Example 1. Resolve into factors   7 x2 - 19x – 6.

First trial, write down (7x  3)  (x  2) noticing that 3 and 2 must have opposite signs. These factors give 7 x2, and – 6 for the first and third terms but since 7 X 2 – 3 X 1 = 11, the combination fails to give the correct coefficient of the middle term. Next try, (7x, 2) (x, 3). Since 7 X 3 – 2 X 1 = 19 these factors will be correct. If we insert the signs so that the negative shall become predominate. Thus 7 x2 - 19x – 6 
= 7 x2 – 21x + 2x – 6
= 7x (x – 3) + 2 (x – 3)
= (x – 3) (7x + 2)
 
Example 2. Resolve into factors  x2 – 3 x - 54.
In the given equation the third term is negative. The second terms of the factors must be such that their product us – 54 and their algebraic sum - 3. Hence they must have opposite signs, and the greater of them must be negative in order to give its sign to their sum

    x2 - 3x – 54
 = x2 - 9x + 6x – 54       
             Or
 = x (x - 9) + 6 (x - 9)
 = x2 - 3x – 54 
 = (x - 9) (x + 6)

Example 3. Resolve into factors  x2 - 10x + 24.
The second term of the factors must be such that their product is + 24 and their sum -10, it is clear that they must be - 6 and - 4.
 
   x2 - 10x + 24
= x2 - 6 x - 4 x + 24
= x (x - 6) - 4 (x - 6)
             Or  
x2 - 10x + 24 = (x - 6) (x - 4)

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