Simple Trinomials as Products of Binomials
Sometimes algebraic expressions of the form
ax^{ 2} + bx + c (1)
where a, b, and c stand for actual numbers, can be written as
a product of two binomials. This is considered a worthwhile
property primarily when the original coefficients a, b, and c,
and any numbers occurring in the resulting binomial factors are
all whole numbers. When this situation occurs, it gives us a way
of factoring such trinomials, rewriting them as a product of two
somewhat simpler expressions.
Notice that
(x + a)(x + b) = x^{ 2} + abx + ab
Thus, when we are given trinomial of the form
x^{ 2} + dx + e
where the coefficient of x^{ 2} is 1, we may be able
to achieve this kind of factorization if we can find two whole
numbers, a and b, such that
a + b = d,
the coefficient of x in the original trinomial, and,
ab = e,
the constant term in the trinomial.
(You might think that the problem is now quite simple, since
we have two unknowns, a and b, and two equations, and so
it’s just a matter of solving the system of equations.
Unfortunately, there are two problems with this: most
importantly, we are only interested in solutions which are whole
number values, and also, the second equation here is not a linear
equation. This doesn’t mean you can’t solve the system
of equations given above in specific instances. In the example
below, we will demonstrate a systematic inspection method which
produces the values a and b required here when they actually
exist.)
Example 1:
Factor x^{ 2} + 5x + 6 if possible.
solution:
A minute’s examination indicates that these three terms
contain no common monomial factors. However, since this
expression has the pattern of expression (1) above with the
coefficient of x^{ 2} equal to 1, there is the
possibility of factoring this expression into the product of two
binomials, (x + a)(x + b). For this to be possible, we need to
find two numbers, a and b, such that
a + b = 5
and
ab = 6.
We could just sit for a while and wait for inspiration, or we
could test out a few guesses. However, we can be a bit more
systematic. The more restrictive condition is that ab = 6. There
aren’t many possibilities. We’ll make a little table
listing all of the pairs of whole numbers whose product is 6:
a 
b 
a + b 
1 
6 
7 
2 
3 
5 
1 
6 
7 
2 
3 
5 
(You might think that a = 6, b = 1, for instance, should also
be in our table. However, swapping values of a and b just amounts
to reversing the order of the factors in the final product, and
so does not give any new possibilities. It is important to
consider both positive and negative factors which multiply to
give +6 in this case, however.)
From this table, we see that there are only four pairs of
whole numbers which multiply to give +6. We’ve also listed
the sum of the two numbers in each case, and you can see that one
of these pairs of values does sum to +5. Thus, a = 2 and b = 3
seems to satisfy the requirements here. Checking,
(x + 2) (x + 3) = (x + 2)(x) + (x + 2)(3)
= x(x + 2) + 3(x + 2)
= x^{ 2} + 2x + 3x + 6
= x^{ 2} + 5x + 6
which is identical to the original expression in this example.
Hence, in factored form we can write
x^{ 2} + 5x + 6 = (x + 2) (x + 3)
(Note that if the table above containing pairs of whole
numbers with a product of +6 had contained no row in which the
two numbers summed to +5, this would be proof that the type of
factorization we were attempting could not be achieved. Thus,
this systematic inspection method provides a factorization of a
trinomial like this when such a factorization exists, and
demonstrates that no such factorization is possible when that is
the case.)
