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Simple Trinomials as Products of Binomials

The Most General Trinomial

Both methods work only for trinomials in which the coefficient of the x 2 term is 1. More general trinomials can sometimes also be factored. Note that

(ax + b)(cd + d) = ac · x 2 + (ad + bc)x + bd

So, a general trinomial (one in which the coefficient of x 2 is a whole number different from 1) can be factored into the form of a product of two binomials, (ax + b)(cx + d), if we can find four whole numbers, a, b, c, and d, satisfying

ac = the coefficient of x 2

ad + bc = the coefficient of x


bd = the constant term

Analyzing these conditions can require quite a lengthy process, depending on the specific values of the various coefficients, because of the large number of possibilities that must be examined, and the complexity of the second condition.

The easiest approach is to make a table of sets of whole number values of a, b, c, and d, which satisfy the first and third conditions above. These are the most restrictive conditions, and so will narrow down the possibilities to be examined as much as possible. For each quadruple of values in the table, the value of ad + bc would then be calculated to determine which quadruple, if any of the possibilities, satisfies the second condition. If such a set of values is found, the factorization follows according to the pattern shown in the box above. If no such set of values is found, the trinomial cannot be factored in this way.

We illustrate this method with a simple example.

Example 1:

Factor 6x 2 + 13x – 5 as much as possible.


The three terms here contain no common monomial factors, so the only possibility is to try to factor this expression into the product of two binomials. Since the coefficient of x 2 is not equal to 1, we must seek a factorization of the form

(ax + b)(cx + d).

The numbers in this product must satisfy

ac = 6 (coefficient of x 2 )

bd = -5 (constant term)

ad + bc = 13 (coefficient of x).

Since ac = 6 is a positive value, we know that a and c must have the same signs. The positivevalued possibilities for a and c are

a = 1 and c = 6

a = 6 and c = 1

a = 2 and c = 3

a = 3 and c = 2

We don’t need to consider negative values of a and c as long as we include both positive and negative possibilities for b and d. The list of four possible pairs of values for a and c above might appear at first to duplicate each actual possibility. However, since a occurs in the factor with b, and c occurs in the factor with d, the situation a = 1 and c = 6 is actually different than the situation with a = 6 and c = 1.

The condition bd = -5 restricts possible values of b and d to the combinations

b = 1 and d = -5

b = -1 and d = 5

b = 5 and d = -1


b = -5 and d = 1.

Note that our list includes the positives and negatives of all values of b which divide evenly into -5. This is necessary to ensure that our analysis includes all possible combinations of whole numbers that might work here.

The four possible pairs of values for a and c can now be matched with each of the four possible pairs of values for b and d, giving a total of 16 possibilities to be check. (You can see that when the coefficient of x 2 and the constant term of the trinomial have a lot of whole number factors, the combined number of possibilities to be examined at this stage can become very large, so that this factorization can become a very tedious job – and often the end result is the conclusion that the trinomial cannot be factored!)

Anyway, setting the analysis up in a table gives:

Thus, it appears that

6x 2 + 13x – 5 = (2x + 5)(3x – 1)


(2x + 5)(3x – 1) = (2x + 5)(3x) + (2x + 5)(-1)

= (3x)(2x + 5) +(-1)(2x + 5)

= (3x)(2x) + (3x)(5) + (-1)(2x) + (-1)(5)

= 6x 2 + 15x – 2x 5

= 6x 2 + 13x – 5

as required. Our proposed factorization of the original trinomial has been verified.

Two, no three, final comments with regard to the method demonstrated with the example above:

(i) It is logically ok to sort of randomly test various possibilities in hope of finding an acceptable combination of values that might work without having to test all of the possibilities in the table. It is important to set the table up systematically however. If you just try random combinations of values of a, b, c, and d, without some way of keeping track of which combinations you’ve tested, you’ll end up going in circles, almost certainly. Of course, the only way to conclude that a trinomial of this type cannot be factored is to make sure you test all possible combinations of values of these four constants. (Yes, if you know how to solve a quadratic equation, you can achieve this factorization or demonstrate that no factorization is possible with quite a lot less work.)

(ii) You can see from the table of test cases above that we still have duplication. Try to figure out how to avoid at least some of the duplication that occurred above. (Hint: the product (ax + b)(cx + d) is identical to the product (cx + d)(ax + b).)

(iii) This is really a very tedious type of problem to solve even under the best of circumstances. There are more powerful and efficient methods for finding the factors of a trinomial which get around the need to enumerate large sets of possible factorizations, but these methods are also more complicated and beyond the scope of this lesson.

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