Consider the expression 3x^{2}
+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,

3xNow, factor by grouping:^{2}+ 14x + 8 =

3x^{2}+ 2x + 12x + 8

3x^{2}+ 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)/xNow, check the TABLE in the Y0 column. You are looking for 14, the coefficient of the middle term.

Y0: (3*8)/x + x

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)/xThen, store the "a", "b", and "c" coefficients in A, B, and C, respectively.For example, key: 3 sto A ENTER

Y0: Y9 + x = B

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.