Simple Trinomials as Products of Binomials
Examples with solutions
Example 1:
Factor x^{ 2} + 10x + 25 as much as possible.
solution:
The constant term here is a perfect square: 25 = 5^{ 2}
. Its square root is Â± 5. But 2 Â· x Â· 5 = 10x, which happens
to be the coefficient of x. So, we can write
x^{ 2} + 10x + 25 = (x + 5)^{ 2}
You can check that this is correct by multiplying the
righthand side to remove the brackets, to confirm that the
result obtained is the expression on the lefthand side.
Example 2:
Factor x^{ 2}  10x + 25 as much as possible.
solution:
This expression is very similar to the one in Example 1. The
constant term is a perfect square, with square root of Â± 5. In
this case, the coefficient of the x term is 10 = 2 x (5), so
the trinomial has the form in the box above, but with a = 5.
Thus, we conclude that
x^{ 2 } 10x + 25 = (x – 5)^{ 2}
This is easily verified:
(x – 5)^{ 2} = (x – 5)(x – 5)
= x(x – 5) + (–5)(x – 5)
= x^{ 2 } 5x 5x + (–5)^{ 2}
= x^{ 2 } 10x + 25
as required.
Example 3:
Factor x^{ 2 } 10x  25 as much as possible.
solution:
At first glance, this trinomial seems to have the same form as
the trinomials in both Examples 1 and 2. However, matching this
trinomial with the pattern shown in the box just above is not
correct. Notice that in the template formula above, all terms are
connected by ‘+’ signs. Thus, to match the trinomial in
this example with that template, we need to write
x^{ 2} 10x  25 = x^{ 2} 10x + ( 25)
Now we see that the constant term must be considered to be
25, not 25, and so it is not a perfect square. Thus, this
trinomial cannot possibly match the pattern to potentially be
equivalent to the square of a binomial.
You could still try to factor this trinomial using the more
general method, to see if it can be factored into a product of
the form (x + a)(x + b). The familiar table of possible values of
a and b is:
a 
b 
a + b 
1 
25 
24 
5 
5 
0 
1 
25 
24 
5 
5 
0 
Since none of the pairs of values which multiply to give 25
also sum to give 10, we conclude that no factorization of any
form of this trinomial is possible.
Example 4:
Factor x^{ 2} 10x + 25 as much as possible. Use the
general method for factoring trinomials.
solution:
This is the same trinomial as considered in Example 2 above.
In this case, we will not make use of the special form of the
coefficients to recognize that it is equivalent to the square of
a binomial. Instead, we’ll confirm that the general method
for factoring simple trinomials will automatically generate that
result.
So, if this trinomial is factorable, it will be into the form
(x + a)(x + b), where
a + b = 10
and
ab = 25
The possible pairs of whole numbers which multiply to 25 are
listed in the table
a 
b 
a + b 
1 
25 
26 
5 
5 
10 
1 
25 
26 
5 
5 
10 
The fourth row of this table gives a sum of 10. Thus using a
= 5 and b = 5 (that is, a = b = 5), we get
x^{ 2}  10x + 25 = (x + (5)) (x + (5)) = (x of the
above – 5)(x – 5) = (x – 5)^{ 2}
as before.
