2. Factoring By Grouping Example:
Factor 𝑥2 − 2𝑥 + 𝑥𝑦 − 2𝑦 completely
Solution:
𝑥2 − 2𝑥 + 𝑥𝑦 − 2𝑦 = 𝑥2 − 2𝑥 Group the terms
= 𝑥 𝑥 − 2 + 𝑦 𝑥 − 2 Factor out x and y
= 𝑥 − 2 𝑥 + 𝑦 Factor.
Some Polynomials may contain
common monomial or binomial
factors. When a polynomial has
four or more terms, common
factors can sometimes be used in
factoring by grouping. The terms
are grouped such that the common
factor in a group is also the
common factor in the other group.
4. Factoring by Grouping and
Factoring General Trinomial
Example:
Factor 4𝑥𝑦 − 𝑥𝑧 + 4𝑥𝑤 − 𝑦𝑧 completely
Solution:
4𝑥𝑦 − 𝑤𝑧 + 4𝑥𝑤 − 𝑦𝑧
= 4𝑥𝑦 + 4𝑥𝑤 − 𝑦𝑧 + 𝑤𝑧 rearrange the
terms; 4𝑥𝑦 and 𝑤𝑧 have no common
factor.
= 4𝑥 𝑦 = 𝑤 − 𝑧(𝑦 + 𝑤)
At times, the basic approach to be
used to factor can be recognized
only after rearranging and
regrouping the terms
5. Factoring by Grouping and
Factoring General Trinomial
Example:
Factor 4𝑥𝑦 − 𝑥𝑧 + 4𝑥𝑤 − 𝑦𝑧 completely
Solution:
4𝑥𝑦 − 𝑤𝑧 + 4𝑥𝑤 − 𝑦𝑧
= 4𝑥𝑦 + 4𝑥𝑤 − 𝑦𝑧 + 𝑤𝑧 rearrange the
terms; 4𝑥𝑦 and 𝑤𝑧 have no common
factor.
= 4𝑥 𝑦 = 𝑤 − 𝑧(𝑦 + 𝑤)
6. Use these steps when factoring four or more terms by grouping:
1. Rearrange and group terms. Collect the terms into two groups so that each group has a
common factor.
2. Factor within a group. Remove the common factor from each group.
3. Factor the entire polynomial. Remove the common binomial factor from the entire
polynomial.
4. Write the expression as a product of the common factor and the remaining factor.
7. Factoring 𝑥2
+ 𝑏𝑥 + 𝑐
Example:
Factor 𝑥2
+ 5𝑥_6 completely.
Solution: 𝑥2
+ 5𝑥 + 6
The factors are of the form (𝑥+ )(𝑥+ ). We
need two numbers whose product is 6 and whose
sum is 5.
To check your answer, use the FOIL method to
multiply the factors. 𝑥 + 2 𝑥 + 3 = 𝑥2 + 3𝑥 + 2𝑥 + 6
= 𝑥2 + 5𝑥 + 6
Factoring a trinomial of the form
𝑥2 + 𝑏𝑥 + 𝑐 means to express the
trinomial as the product of two
binomials of the form 𝑥 + 𝑚 ( 𝑥 +
Factors of 6 Product Sum
1,6 (1)(6)=6 1+6=7
2,3 (2)(3)=6 2+3=5
8. Factoring by Grouping 𝑎𝑥2
+ 𝑏𝑥 +
𝑐 𝑤ℎ𝑒𝑟𝑒 𝑎 ≠ 1
Example:
Factor 3𝑥2
+ 5𝑥 + 2
Solution:
Method 1:Trial and Error
a. Factor the first and last terms. 3𝑥2
→ 3𝑥 ∙ 𝑥
2 → 2 ∙ 1
b. Write the possible pairs of factor. 3𝑥 + 2 𝑥 + 1
(3𝑥 + 1)(𝑥 + 2)
c. Check for the correct middle term using FOIL method.
3𝑥 + 1 𝑥 + 2 ⇒ 6𝑥 + 𝑥 = 7𝑥 Wrong middle term
3𝑥 + 2 𝑥 + 1 ⇒ 3𝑥 + 2𝑥 = 5𝑥 Correct middle term
d. Answer: (3𝑥 + 2)(𝑥 + 1)
When the coefficient 𝑎 of the term
𝑎𝑥2
is not 1, there are more
possibilities for factoring to
consider. We will consider two
methods of factoring this trinomial
using trial and error, and factoring
by grouping.
9. Method 2:Using Factoring By
Grouping
3𝑥2 + 5𝑥 + 2
a. Get the product of the first and last terms.
3𝑥2 2 = 6𝑥2
b. Find two numbers whose products is 6𝑥2
and
whose sum is 5𝑥.
The numbers are 2𝑥 𝑎𝑛𝑑 3𝑥.
c. Arrange the terms and apply factoring by grouping.
3𝑥2 + 5𝑥 + 2 = 3𝑥2 + 2𝑥 + 3𝑥 + 2
= 3𝑥2
+ 2𝑥 + 3𝑥 + 2
= 𝑥 3𝑥 + 2 + 1 3𝑥 + 2
= (3𝑥 + 2)(𝑥 + 1)
Factors of 𝟔𝒙 𝟐 Product Sum
6𝑥, 𝑥 6𝑥 𝑥 = 6𝑥2
6𝑥 + 𝑥 = 7𝑥
2𝑥, 3𝑥 2𝑥 3𝑥 = 6𝑥2
2𝑥 + 3𝑥 = 5𝑥
10. Here are the steps in factoring a trinomial in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐, where 𝑎 ≠
1using the trial and error method:
1. Look for a common factor, if any.
2. Factor the first term 𝑎𝑥2
3. Factor the last term c.
4. Look for the factors of 𝑎𝑥2
and such that the sum of their product is the middle term 𝑏𝑥.
The Following are the steps in factoring trinomial in the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐, 𝑤ℎ𝑒𝑟𝑒 𝑎 ≠ 1using
the grouping method:
1. Look for a common factor, if any.
2. Get the product of the first and last terms, then test if it is possible to find factors whose sum is
equal to the middle term.
3. Think of two expressions whose product is equal to (𝑎𝑥2
)(c)and whose sum is equal to 𝑏𝑥.
4. Arrange the terms and apply the basic steps in factoring by grouping.
11. Remember
the steps in factoring by grouping are as follows:
1. Rearrange and group terms.
2. Factor within s group.
3. Factor the entire polynomial.
4. Write the expression as a product of the common factor and remaining factor.
