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Simple Trinomials as Products of Binomials

Examples with solutions

Example 1:

Factor x 2 – 2x – 35 as completely as possible.

solution:

The terms of this expression contain no common monomial factors. However, since this expression has the form of a trinomial in x, with the coefficient of x 2 being 1, factoring into the product of two binomials, (x + a)(x + b) may be possible. We would need to find values for a and b satisfying

a + b = -2

and

ab = -35

As in the previous example, we begin by making a table of all pairs of whole numbers which multiply to give, in this case, -35. The easiest way to make sure our list is complete is to make sure that the value of ‘a’ covers all positive and negative whole numbers which divide evenly into -35. The value of ‘b’ in each case is obvious, because we need to have ab = -35.

It appears that there are eight candidate pairs of whole numbers that multiply to give -35. (However, if you examine the table carefully, you’ll see that only four of these pairs of values are unique. The last four rows essentially duplicate the first four rows, amounting to swapping the order of the two possible factors in the eventual product. So, we’ve done twice as much work here as was really necessary.)

To achieve the proposed factorization, one of these pairs of numbers must sum to -2. From this table, this is seen to occur when a = 5 and b = -7. Thus, it appears that

x 2 – 2x – 35 = (x + 5)(x – 7)

is the required answer.

Checking,

(x + 5)(x – 7) = x(x – 7) + 5(x – 7)

= x 2 -7x + 5x – 35

= x 2 - 2x – 35

confirming our answer.

 

Example 2:

Factor as completely as possible: x 2 + 8x + 10.

solution:

There are no common monomial factors here, but the expression has the form of a trinomial of the type considered in the previous two examples. Thus, a possible factorization has the form of a product of two simple binomials, (x + a)(x + b), if we can find two numbers, a and b, satisfying

a + b = 8

and

ab = 10

Setting up the usual table of pairs of whole numbers with a product of +10, we get

(Here, we listed values of ‘a’ which divide evenly into +10, only for |a| < |b|, realizing from the previous example that to also include rows in the table for a = ± 5 and a = ± 10 will just duplicate the information in these first four rows.)

To get this factorization to work, we need to find a pair of these whole numbers which sum to 8. However, the table contains all possibilities, and none of the rows give a sum of 8. So, we must conclude that the given trinomial cannot be factored.

In the special situation that a = b, we get the form

(x + a)(x + a) = (x + a) 2 = x 2 + 2ax + a 2

Thus, a trinomial in which the coefficient of x 2 is 1 is the square of a binomial if

(i) the constant term is a perfect square (of a whole number)

and

(ii) the coefficient of the x term is double the square root of the constant term.

You don’t really need to memorize this formula as a special case, because the more general method described above will also work in this case. You’ll just find that the two whole numbers, a and b, that you get from the analysis will be equal.

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