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 Â· 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
and
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.
solution:
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
and
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)
Checking,
(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.
