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.
3x2 + 14x + 8 =Now, factor by grouping:
3x2 + 2x + 12x + 8
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.
2nd TBLSET (delta)Tbl = 1
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
Method 2 (probably better)
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
To repeat for another problem, merely store new values in A, B, and C.