![]() |
Teacher2Teacher |
Q&A #104 |

T2T || FAQ || Ask T2T || Teachers' Lounge || Browse || Search || Thanks || About T2T

|
View entire discussion [<<prev] [next>>]
Actually, the most effective way I have found is to use the
TIC-TAC-TOE method.
Create a 9 square board.
a---c---ac Row one takes the a coefficient in square 1, c in the second
square and the product ac in the third.
Next do column 3. What do you multiply to get ac but add to
equal "b"? (Note that we haven't used b yet). Place those
factors along column 3 in rows 2 and 3.
Let me take you through an example: 6x^2-7x-3
6 -3 -18 What multiplies to equal -18 and adds to equal = -7 ?
-9 and +2. Put them in position.
6 -3 -18
-9
2 (Either order is fine.) Now, what are two factors of -9,
such that one divides evenly into the 6 located in row 1 and
the other into the -3, also in row 1?
3 and -3. Position them.
6 -3 -18
3 -3 -9 You can repeat the same thing for factors of 2 or just
2 1 2 divide the 3 into 6 and get 2, -3 into -3 and get +1.
Now look at the square in the lower left
3 -3
2 1 The diagonals of these form the factored trinomial
(3x+1)(2x-3).
It is somewhat difficult to illustrate this on the computer... I hope this
makes sense!
Post a public
discussion message |
[Privacy Policy] [Terms of Use]

Math Forum Home ||
The Math Library ||
Quick Reference ||
Math Forum Search

The Math Forum is a research and educational enterprise of the Drexel University School of Education.