Unit 2: Topic 4: Factoring Trinomials

Competencies

·   Identify the terms factor, trinomial,

·   Factorize trinomials

Suggested ways of teaching this topic: Teacher’s explanation with guided practice

Starter Activities

The process of factoring is essentially the opposite of the FOIL Method, which is a process used to multiply two binomials. At the beginning, the teacher shall make sure the students understand the FOIL Method.

The teacher could ask students to examine the following expression, in pairs or small groups, which consists of one binomial in parentheses multiplying another binomial in parentheses.

(2k + 7)(3k - 10)

Let students use the FOIL Method to simplify the expression.

Expected answer is the work below and a final result which is a trinomial.

6k² - 20k + 21k - 70
6k² + k - 70

The teacher shall make students notice that a trinomial consists of three terms.

Lesson Notes

Then the teacher should ask students “how can we go from the trinomial to factors or factor a trinomial like this into two binomials?”

The first problem could be of the following type:

m² + 10m +16

Before attempting to factor any more, there are a few simple questions you can ask to make sure that the expression is factorable as a trinomial.

 

Question	Answer and Reason
After the GCF has already been factored out, are all variables in the first (m2) and last (16) terms to an even power?	Yes. (m2 is the only variable in the first and last terms, and its exponent, 2, is an even number.)
Are all of the middle term's (10m) exponents to half of the power of one of the exponents on the outside terms?	Yes. (In 10m, the middle term, m is the only variable. Its exponent is not shown and therefore is 1 which is half of 2, the exponent of m in the first term.)

The next step is to write sets of open parentheses, side by side.

(          )(          )

Handling Variables

On the left side of each set of parentheses, write all of the variables from the first term of the trinomial with half their exponents. The first term is m², thus after the exponent is cut in half, m is placed inside each set of parentheses.

(m      )(m      )

Since all of the terms in the original set of parentheses are positive, two plus signs are placed below.

(m +    )(m +    )

Handling Coefficients

Now identify the coefficient of the first and last terms in the original set of parentheses, m² + 10m + 16. The coefficients are 1 and 16. Now write all pairs of factors of 1 in a vertical column and then write all pairs of factors of 16 in another vertical column.

Factors of 1 1 * 1

Factors of 16 1 * 16 2 * 8 4 * 4

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. But how is this done? For the number 1, there is only 1 pair of factors, eliminating any choice. Thus, the 1s can be inserted on the left side of each set of parentheses.

(1m +    )(1m +    )

Finally, a pair of factors of 16 must be chosen. This is a matter of trial and error. To make sure all pairs are considered, start using factors at the top of the list then check each pair below it, in order.

We start by choosing the first pair, 1 * 16. We place the factors on the right side of each set of parentheses.

(1m +  1 )(1m + 16)

Now test whether this is the correct pair. Multiply the expression using the FOIL Method:

(1m² + 16m + m + 16)
(1m² + 17m + 16)

Note that the resulting trinomial is not the same as the one we started with, so this is the incorrect pair of factors of 16. Try the next pair:

(m +  2 )(m +  8 )
(m² + 8m + 2m + 16)
(m² + 10m + 16)

Since the result here is equivalent to the original, the correct pair of factors of 16 has been chosen. Additionally, the answer to the problem is: (m + 2)(m + 8)

Factoring a Trinomial with a Negative Sign

Example 1: 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 the variable 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.

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

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

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

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

The expression s² - 5s + 6 is equivalent to the original expression, therefore (s - 2)(s - 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.

Example 2: Factor 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.

(p        )(p        )

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.

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

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

Sum and Product Method

Example 3: Factor   6c² + c - 12

The expression is a trinomial, like the expressions which we factored on previous pages. There is a small variation in this problem; however, the first term has a coefficient that is not 1.

This expression will be factored much like the others, the main difference is that we will need to find the correct pair of factors for the first term's coefficient (6) in addition to the correct pair of factors for the last term's coefficient (-12).

Thus, try to find two numbers whose product is -72 and whose sum is 1.

To do this, we need to write possible factors of -72.

Factors of -72 1 * -72 -1 * 72 2 * -36 -2 * 36 3 * -24 -3 * 24 4 * -18 -4 * 18 6 * -12 -6 * 12 8 * -9 -8 * 9

The last pair satisfies our condition, sum = 1 and product = -72.

Thus, the middle term of the trinomial can be written as -8c +9c while the leading term as (2c)(3c) and the last term as (3)(4).

We can now have 6c² + c – 12 =   [(2c)(3c)– 8c]+[9c –(3)(4)]

Taking 2c a common term the first sum and 3 from the second we can have:

2c (3c – 4) + 3(3c – 4).

This is same as distributing (3c – 4) in to (2c + 3).

Thus, the factors are (3c – 4) and (2c + 3).

Example 4: Factorize -5x² + 4x + 1

Solution:

Start by taking out the minus sign from the leading term.

We get:- ( 5x² – 4x – 1)

Then, leaving the minus sign as it is we can factor the trinomial in the bracket.

Here, we need to find two numbers whose product is (5)(-1)= -5 and whose sum is –4.

The teacher may ask: “Which do you think are those numbers?” ‘why?” to make sure that students know what they are doing.

Expected Answer: -5 and 1.

Then we can write the given trinomial as follows:

-5x² + 4x + 1 = -( 5x² – 4x – 1) = - (5x.x -5x + x – 1)

= - [5x(x – 1) + 1(x-1)]

= -(5x + 1)(x-1)

Thus, the factors -(5x + 1) and (x-1).

Remember students should be reminded about the trinomials that can’t be factored using sum and product rule. Some examples are 2x² – 4x + 5; x²– x + 1;

Concluding Activities

The teacher has to make sure that the students have understood the process of factoring. To factor trinomials of the form ax² +bx + c, we need to find two numbers whose sum is b and whose product is a•c. If we can find the pair, say p and q, we can write the middle term as a sum of px+qx. Then, what remains is taking out common and writing the factors.

Practice Exercises

1.      Find the products

a)      (2x + 7)(5x + 1)

b)      (3x – 1)(2x+3)

c)      (2 – 5x)(x – 1)

d)     – (x –3)(3x +1)

2.      Factor the following trinomials

a)      x² – 9

b)      3x² – 1

c)      x² – 6x + 9

d)     x² + 4x – 5

e)      3x² – 4x + 3

f)       8 – 2c – c²

g)      – 5m – m² – 4