Factoring Trinomials using the "AC" or Grouping Method


 

Consider the expression 3x2 +14x + 8 for factoring, where a=3, b=14, and c=8.

First, multiply the coefficients "a" and "c": 3 * 8 = 24.

Now, we are looking for a pair of integer factors whose product is 24 and whose sum is 14, which is  the "b" coefficient.

We find that 2 * 12 = 24 and 2 + 12 = 14. Therefore, 2 and 12 are the numbers we want.

Now, replace the middle term of the trinomial, 14x, with two equivalent terms, 2x +12x.
Therefore,

3x2 + 14x + 8 =
3x2 + 2x + 12x + 8
Now, factor by grouping:
3x2 + 2x + 12x + 8 =
x(3x + 2) + 4(3x + 2) =
(3x + 2)(x + 4)

To use the TI-82/83 to help find the necessary replacements for 14, first make sure the table you
will be creating increments by one.
Key:
            2nd TBLSET (delta)Tbl = 1
Then,

Method 1
Key:

Y9: (3*8)/x
Y0: (3*8)/x + x
Now, check the TABLE in the Y0 column. You are looking for 14, the coefficient of the middle term.
When you find it, look across the same row. The numbers 2 and 12 will be in the X column and the Y9 column.
 

Method 2 (probably better)
Key:

Y9: (A*C)/x
Y0: Y9 + x = B
Then, store the "a", "b", and "c" coefficients in A, B, and C, respectively.For example, key: 3 sto A ENTER

Now, check the TABLE in the Y0 column. You are looking for a 1.The other values should be 0 (or ERROR).
When you find the 1, look across the same row. The numbers 2 and 12 will be in the X column and the Y9
column.

 

To repeat for another problem, merely store new values in A, B, and C.