Consider a polynomial expression of the form ax2 + bx + c.
The Intermediate Algebra book has a discussion of the ac method that makes it much worse than it has to be.
It sounds simple enough ...
The polynomial can be factored if there are two factors of ac whose sum is b.
... but then the text makes it very time intensive to find the solution.
The way I teach it is much easier. Ignore negatives until the very end.
There are two main situations. One where the constant is positive, and one when the constant is negative.
ax2 + bx + cor ax2 + bx - c
The polynomial can be factored only if there are two factors of ac that add to be the absolute value of b.
The polynomial can be factored only if there are two factors of ac that have a difference of the absolute value of b.
If you ignore the sign of the middle number so that we don't have to keep saying the absolute value of b ...
Are there two factors of ___________ ( a times c )
whose _____________ ("sum" or "difference" depending on c)
is _______ (b without the sign)?
If you can answer yes - then it can be factored and the two factors you found which worked in answering the question will work in the factoring also.
Question: Are there two factors of 2(6) = 12 whose sum (because the last # is positive) is 7 (middle number)?
Answer: Yes, 4 and 3. So this problem will factor.
Rewrite the original problem and factor by grouping.
| Expression | Comments |
|---|---|
| 2 + 7a + 6a2 | Original expression |
| 2 + 4a + 3a + 6a2 | Rewrite the 7a as 4a + 3a, putting the largest value first and using the same sign as the original middle value |
| 2 (1 + 2a) + 3a (1 + 2a) | Factor by grouping. |
| ( 2 + 3a ) ( 1 + 2a ) | The (1 + 2a) is a common factor for both terms |
Question: Are there two factors of 2(15) = 30 whose difference (because the last # is negative) is 7 (middle number)?
Answer: Yes, 10 and 3.
Rewrite the original problem and factor by grouping.
| Expression | Comments |
|---|---|
| 2x2 + 7x - 15 | Original expression |
| 2x2 + 10x - 3x - 15 | Rewrite the 7x as 10x - 3x, putting the largest value first and using the same sign as the original middle value |
| 2x (x + 5) - 3(x + 5) | Factor by grouping. |
| (2x - 3) (x + 5) | The (x + 5) is a common factor for both terms |
Notice how there is a common factor between the two terms after grouping the first two together and the last two together.This is NOT A COINCIDENCE! If you can answer yes to the question, it will factor in this method.
Question: Are there two factors of 3(4) = 12 whose sum (because the last # is positive) is 5 (middle number - ignore the sign)?
Factors of 12 are ...
Answer: NO! The problem won't factor, write "prime" and go on.
Factor out the greatest common factor of 200 first to get 200 ( 3 - 4t - 4t2 )
Question: Are there two factors of 3(4) = 12 whose difference (because the last # is negative) is 4 (middle number - ignore the sign)?
Answer: Yes, 6 and 2.
| Expression | Comments |
|---|---|
| 200 ( 3 - 4t - 4t2 ) | Original expression |
| 200 ( 3 - 6t + 2t - 4t2 ) | Rewrite the -4t as -6t + 2t, putting the largest value first and using the same sign as the original middle value |
| 200 [ 3 (1 - 2t) + 2t (1 - 2t) ] | Factor by grouping. |
| 200 ( 3 + 2t )( 1 - 2t ) | The (1 - 2t) is a common factor for both terms |