# 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
right-hand side to remove the brackets, to confirm that the
result obtained is the expression on the left-hand 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. |