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# 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.