Splitting the Middle Term

The challenge is to factor a trinomial of the form ax² ± bx ± c where a, b, and c are integers and a>0. The method for doing this is given below. This next part gives the reasoning behind the method.

Justification

We will start by making a few observations assuming that the correct factorization looks like (dx ± e)(fx ± h). First, the only way that the sign before c can be negative is if we are multiplying a sum by a difference as in (dx + e)(fx - g) or (dx - e)(fx + g) and in either case we have a sum and a difference. Second, if we are multiplying a sum times a sum, as in (dx + e)(fx + g) then we must be adding the three terms as in ax² + bx + c Third, if the sign before the c is positive but the sign before the b is negative, then the correct factorization must be the product of two differences as in (dx - e)(fx - g) We will expand each of the three cases:

product of sums	product of a sum and a difference	product of differences
(dx + e)(fx + g)	(dx + e)(fx - g)	(dx - e)(fx - g)
(dx + e)(fx) + (dx + e)(g)	(dx + e)(fx) - (dx + e)(g)	(dx - e)(fx) - (dx - e)(g)
(dx)(fx) + (e)(fx) + (dx)(g) + (e)(g)	(dx)(fx) + (e)(fx) - ( (dx)(g) + (e)(g) )	(dx)(fx) - (e)(fx) - ( (dx)(g) + (e)(g) )
dfx² + efx + dgx + eg	dfx² + efx -( dgx + eg )	dfx² - efx - ( dgx - eg )
dfx² + (ef + dg)x + eg	dfx² + efx - dgx - eg	dfx² - efx - dgx + eg
	dfx² + (ef - dg)x - eg	dfx² - (ef + dg)x + eg
All three signs are positive and the middle term is the sum of ef and dg.	The sign before the last value is negative and the middle term is the difference between ef and dg.	The sign before the last calue is positive and the middle term, though preceded by a subtraction, is the sum of ef and dg.
In all cases, a must be equal to df and c must be equal to eg.

If we multiply the values of a and c we get the same answer as when we multiply df and eg. So,

ac	=	(df)(eg)
	=	defg
	=	(ef)(dg)

Thus, the middle term of our trinomial, ax² ± bx ± c, must be the sum or difference of two factors of the value ac. Therefore, if we look at all of the factor pairs of ac we should find a pair that has their sum or difference [choosing the appropriate one for the signs of the trinomial] equal to the middle coefficient, b, in our trinomial. If we find such a situation, then we can substitute the sum or difference for that middle coefficient and our factorization will move forward from there.

Method

The actual method involves finding the factor pairs for the product of a and c. Using the example 6x² + 23x + 20 we multiply 6 and 20 to get 120. Then we find the factor pairs for 120, namely,

1	and	120
2	and	60
3	and	40
4	and	30
5	and	24
6	and	20
8	and	15
10	and	12

Since the original problem has no negative signs, we are looking for a pair that adds to give the middle term, 23. Inspecting the possibilities we see tht 8 and 15 work. Therefore we split the middle term, 23x, into 8x and 15x. Now our problem looks like 6x² + 8x + 15x + 20 We group the first two and the last two terms: (6x² + 8x) + (15x + 20) factor each if possible: 2x( 3x + 4) + 5( 3x + 4) notice the common factor in each, in this case, (3x+4), and factor that out to give (2x + 5)(3x + 4)

A second example is

6x² + 19x - 20 Again, we need to find the factor pairs of 6*20, or 120. We have them above. Since the third term in the problem is preceded by a minus sign we are looking for factor pairs that have a difference of 19. Looking at the factor pairs we see that 5 and 24 work. Split the middle term using 5x and 24x to give 6x² + 24x - 5x - 20 We group the first two and the last two terms: (6x² + 24x) - (5x + 20) factor each if possible: 6x(x + 4) - 5(x + 4) notice the common factor in each, in this case, (3x+4), and factor that out to give (6x - 5)(x + 4) A third example is 6x² -43x + 20 Again, we need to find the factor pairs of 6*20, or 120. We have them above. In this example, the final term is added but the middle term has a subtraction sign before it. Thus, we again look for a factor pair that adds to the middle term, 43. We see that 3 and 40 work. We replace the middle term with these two giving 6x² -3x - 40x + 20 We group the first two and the last two terms: (6x² -3x) - (40x - 20) factor each if possible: 3x(2x - 1) - 20(2x - 1) notice the common factor in each, in this case, (2x-1), and factor that out to give (3x - 20)(2x - 1)

This method works to factor trinomials with integer coefficients, if factoring is possible. It does rely on being able to produce the factor pairs for a value. This is something that a programmable calculator should be able to do. You will find a program to do this at the factor pair page. Of course, if the calculator can find the appropriate factor pair, then how much more is it to ask that the calculator do the factoring. After all, the method outline above is just a step by step solution once you have found the right factor pair. There is a web page demonstrating this kind of a program for your calculator at Factoring a Trinomial. The output on a TI-83 for this kind of program is rather ugly since it just gives the four values for d, e, f, and g [actually, the program gives 5 values with the additional one being a common factor to all three of the original a, b, and c values]. With a little work, one can get the TI-83 to produce slightly better looking output, and this has been done for you on a page named Factoring a Trinomial with pretty output.