Taking the Trial and Error out of Factoring Trinomials

by H. Paul McGuire

San Diego Miramar College

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Phone: (858) 578-3421

4/29/04

Though there are now calculators that will give the factors of certain algebraic expressions, most teachers agree it is still necessary to teach the factoring of trinomials. Textbooks refer to that process as one of “trial and error”. That usually strikes students as strange. “If math is so sure,” they complain, “why is factoring trinomials a bunch of guesswork?” It turns out, there is no guesswork if you use the right formula.

For any trinomial which is quadratic in form, the factors are: 2ax + b (b2 - 4ac). The reason this has such similarity to The Quadratic Formula is because all factorable trinomials are quadratic in form and this formula bridges the gap between The Quadratic Formula and The Factor Theorem, which says that the roots of a quadratic derive from its factors.

For those who care to see the relationship between The Quadratic Formula and The Factor Theorem, here are the steps:

x = -b ( b2 – 4ac)

2ax = -b ( b2 – 4ac)

2ax + b = ( b2 – 4ac)

Then, because  allows for either sign, anytime, then

2ax + b ( b2 – 4ac) = 0
From which, as we will see in the two examples that follow, factors can be derived and the roots of the equation are obtained. The formula applies to a typical trinomial with the following steps:

Given: 8x2 – 2x – 15, then a = 8, b = -2, c = -15

2(8)x +(-2) {(-2)2 – 4(8)(-15)}

16x –2 (4 + 480)

16x – 2 (484)

16x –2  22

(16x – 24)(16x + 20)

Now we throw away any common factors. That leaves:

(2x – 3)(4x + 5)

If this had been the equation 8x2 – 2x – 15 = 0, the roots (also known as solutions or zeros) of the equation would be obtained by solving 2x – 3 = 0 and 4x + 5 = 0. That explains why we throw away common factors: the solutions of 16x – 24 = 0 and 16x + 20 = 0 are the same as those of the “leaner” factors.

Another example:

Given: 6x2 – xy - y2 Here we let a = 6, b = -y, c = -y2

2(6)x + (-y) {(-y)2 – 4(6)(-y2)}

12x – y (y2 + 24y2)

12x –y (25y2)

12x –y  5y

(12x – 6y)(12x + 4y)

Throwing away common factors yields:

(2x – y)(3x + y)

Why then would anyone want to use The Trial and Error Method? Sure-fire as this formula is, using The Trial and Error Method requires less steps than using the formula if we give students a few “considerations” when picking values for the parentheses. The considerations I propose apply at a glance without the tedious process of listing and trying all possible combinations of factors:

1)Be sure the terms are arranged in descending or ascending powers of one variable and have no common factors. If there is a common factor, factor this out first.

2)Select two factors of the first term (variables like x2 or x4 are always split evenly) and if there are multiple possible factors, probability favors the factors closest together. Thus if the first term was 24x2, you would choose 4x and 6x rather than “3x and 8x” or “2x and 12x” or “1x and 24x”. I say probability favors the closest together factors because both in practical applications and in general factoring, the closer factors most commonly work. (If they don’t we cycle through other possibilities.) Put those factors in the front of the two sets of parentheses. For example, if we were factoring 24x2 – 2x –15, we would start the two parentheses as follows: (4x )(6x )

3)Put the two closest together factors of the last term in the back of the two parentheses, but never put common factors in the same set of parentheses. This is because there cannot be a common factor later if there was none in the original problem. (If there had been, we would have taken out any common factors in step one before starting to factor the trinomial.) In the example, we choose 3 and 5 as factors of 15 and put the 5 in with the 6x: (4x 3)(6x 5). Summarizing to this point: The factors of the front term go to the front of each set of parentheses; the factors of the back term go in the back of each set. More concisely: the fronts give the front; the backs give the back.

4)Now the product of the inners and the product of the outers must add together to give the middle term of the original trinomial. Doing this in one’s head is often difficult, so drawing arcs will help.

(4x 3) (6x 5)

18x

20x

We have not yet worried about signs, which seem to be the most difficult for students to juggle in their heads. From here it is easy to see that if we want the middle term to be –2x, obtained from adding 18x and 20x, the 20 must be negative and the 18 positive. If we have drawn the arcs this way, the upper sign now goes to the left parentheses and the lower sign goes to the right. This gives (4x + 3)(6x – 5).

5)Having placed the signs in this way there is just one final “consideration”. Everything else has been rigged to derive from the original trinomial. The fronts give the front, the backs give the back, the inners and outers give the middle term. We need only check whether the signs thus selected give the original back term when multiplied together. In our example, positive 3 and negative 5, when multiplied together, do indeed give the original back term, negative 15. So we are finished. If they had not, we would first start back at step 3 with a different set of factors, and if necessary, return to step 2.

In contrast to the steps required by the formula above, these 5 considerations allow our steps to look like this:

24x2 – 2x – 15

(6x 5)(4x 3)

20x

18x

Now picking the signs on the 20 and 18 and putting the upper sign into the left parentheses and lower sign (the one on 18) into the right set of parentheses happens right on that same step. So the whole factoring is much shorter than using the formula.

One last complication to consider: If the trinomial has a negative sign on the initial term, just factor (-1) out of the trinomial before starting. Thus for:

-8x2 + 2x + 15

We start with -[8x2 – 2x – 15]

Then: -[(4x )(2x )]

I have no guide for where to put 3 and 5 so let’s try -[(4x 3)(2x 5)]

Now with the arcs… -[(4x 3)(2x 5)]

20x

A glance shows there’s no way to get “–2x” from 6x and 20x, so we switch 3 and 5 (consideration 3).

-[(4x 5)(2x 3)]

10x

12x

Now it’s obvious that making 12 negative and 10 positive will yield “-2x”. Applying the “upper sign to left, lower sign to right” principle yields:

-[(4x + 5)(2x – 3)]

Of course we don’t want that negative out front so we distribute it into just one of the parentheses, preferably the one that already has a negative. This gives

(4x + 5)(3 – 2x)

Commuting in the first set of parentheses gives the more “pleasantly” consistent

(5 + 4x)(3 – 2x).

However, all of this could have been avoided by commuting the original trinomial into ascending powers rather than descending ones before factoring:

15 + 2x – 8x2

(5 4x)(3 2x)

This is much simpler. But the “factoring out (-1)” approach would be more desirable if both the initial and the end terms were negative – a very rare situation.

In summary the 5 considerations are:

1)Arrange in descending or ascending powers with no common factors.

2)Fronts give front term, backs give back term.

3)No common factors in any set of parentheses.

4)Draw arcs. Pick signs. Upper to left, lower to right.

5)Check to be sure that, once the signs are inserted, their product gives the sign of the final term. If not, try other combinations.

I hope these easily applied considerations will provide you and your students many happy trinomial factorings. This leaves the Quadratic Formula and The Factor Theorem for more advanced classes. But if students grumble that factoring by trial and error is too hard, you can now offer them a sure-fire alternative (the formula), albeit one which will require more steps.