Home Lessons Calculators Worksheets Resources Feedback Algebra Tutors

## Factoring A Trinomial with a Negative Sign

Examine this expression.

s2 - 5s + 6

Only one of the three terms is negative. As a result, a minus sign should not be factored out as in the previous example. This expression will be factored much like the expression on the previous page, but we will need to work with the minus sign as we build two sets of parentheses.

To start factoring, first write out two empty sets of parentheses.

(            )(            )

### Handling Variables

Now on the left of each set of parentheses, write all of the variables from the first term with half the exponent. The first term is s2, after dividing its exponent by two, we place s on the left of each set of parentheses.

(  s        )(  s        )

The last term in the trinomial does not have any variables, so we do not need to carry any variables into the parentheses as a result.

### Handling Coefficients

Now identify the coefficient of the first and last terms in the original set of parentheses, s2 - 5s + 6. The coefficients are 1 and 6. Now write all pairs of factors of 1 in a vertical column and write all pairs of factors of 6 in another vertical column.

Since this problem involves a negative term, we also include the negative factors of 6.

 Factors of 1 1 * 1 Factors of 6 1 * 6 -1 * -6 2 * 3 -2 * -3

### Choosing Factor Pairs

Now we must choose a pair of factors of 1 and a pair of factors of 16 to insert into the pairs of parentheses. For the number 1, there is only one pair of factors, eliminating any choice. Thus the 1s can be inserted on the left side of each set of parentheses.

(1s        )(1s        )

Now a pair of factors of 6 must be chosen. This is a matter of trial and error. To make sure all pairs are considered, check each pair of factors starting with the pair on the top of the list and working toward the bottom of the list.

A summary of the trial and error process using the FOIL Method is below.

(1s + 1)(1s + 6)
s2 + 6s + s + 6
s2 + 7s + 6

(1s + -1)(1s - 6)
s2 - 6s - s + 6
s2 -7s + 6

(1s + 2)(1s + 3)
s2 + 3s + 2s + 6
s2 + 5s + 6

(1s - 2)(1s - 3)
s2 - 3s - 2s + 6
s2 - 5s + 6

The expression s2 - 5s + 6 is equivialent to the original expression, therefore (1s - 2)(1s - 3) is the final answer.

## An Example With Two Negative Signs

The problem below is similar to the last problem, but it has two negative signs in the expression.

p2 - 20p - 21

First, write out two sets of empty parentheses.

(            )(            )

Write all of the variables from the first term with half the exponent in the front of each set of parentheses. Again, since there are no variables in the last term, nothing is written on the right side of each set of parentheses.

(  p        )(  p        )

Write out the factors of the coefficients of the first and last terms. The previous problems have led to the observation that the first term's coefficient of 1 results in 1 * 1 as the pair of factors chosen.

(1p        )(1p        )

Now the pairs of factors of 21 (both positive and negative) are computed.

Factors of -21
1 * -21
-1 * 21
3 * -7
-3 * 7

Use trial and error to find which pair of factors to use.

(1p + 1)(1p - 21)
p2 - 21p + 1p - 21
p2 - 20p - 21

The expression p2 - 20p - 21 is equivalent to the original expression. Therefore, (1p + 1)(1p - 21) is the answer.

Proceed to the next page for a final example.

 Home Lessons Calculators Worksheets Resources Feedback Algebra Tutors © Copyright 2001-2011 info @ algebrahelp.com