Lesson 1: Overview
Factoring (also known as factorizing) is a mathematical process of breaking down a number or expression into smaller parts (known as factors) that can be multiplied together to get the original number or expression.
For instance, we can factor the number 12 into 3\times 4, where 3 and 4 are known as the factors of 12.
In algebra, there are four techniques to factor an expression
- factor out the greatest common factor (GCF)
- difference of squares
- cross method
- grouping
We should always try to factor out the GCF first.
Next, if the expression has two terms and the two terms are subtracted, we try to factor it using difference of squares.
If the expression has three terms, we try to factor it using the cross method.
Finally, if the expression has more than three terms, we use the “divide and conquer” technique that involves grouping the terms into smaller groups and factoring the separately.
Lesson 2: Factoring out the GCF
To factor out the GCF, we need to first identify the terms in the given expression.
Terms refer to individual elements in an expression or equation that are combined through addition or subtraction. As such, they are separated by +/- signs that are not in a bracket, fraction, or square root sign.
Let’s consider the expression
\begin{aligned} 3x+(2x+5)(x-2)-9 \end{aligned}This expression has one “+” sign and one “-” sign that is not inside any bracket, fraction, or square root sign.
- The “+” sign is between 3x and (2x+5)(x-2), and
- The “-” sign is between (2x+5)(x-2) and 9
Hence, the expression has the following three terms:
- 3x
- (2x+5)(x-2)
- 9
| NOTE |
| Alternatively, we can say that the three terms of the given expression are 3x, (2x+5)(x-2), and -9. In other words, the third term is -9 (negative 9) instead of 9. This is because we can rewrite any subtraction as an addition of a negative term. Specifically, we can rewrite the expression as 3x+(2x+5)(x-2)-9=3x+(2x+5)(x-2)+(-9) Thus, the third term is -9 instead of 9 and it is added to (instead of subtracted from) the second term. Considering the “-” sign in front of 9 as a negative sign (instead of subtraction) is very useful in some cases, such as when we do algebraic expansion, but less so when we do factoring. The key is to remember to add the results. |
After we identify the terms, we determine if there are any common factors among the terms. If there are, we factor out the greatest common factor.
Example 1
Factor out the greatest common factor in the expression 3x^5y+15x^2y^3.
Solution
There are two terms in the expression 3x^5y+15x^2y^3.
- The first term can be broken down into (3)(x)(x)(x)(x)(x)(y).
- The second term can be broken down into (3)(5)(x)(x)(y)(y)(y).
Both terms have a common factor 3.
In addition. the first term has 5 factors of x while the second term has 2. Therefore, two of the xs are common.
The first term also has 1 factor of y while the second term has 3. Therefore, one of the y is common.
As a result, the GCF is 3x^2y.
We can factor out 3x^2y and write the remaining factors inside a pair of parentheses.
For the first term, after factoring out 3, two x‘s, and one y , we are left with \cancel{(3)(x)(x)}(x)(x)(x)\cancel{(y)}=x^3.
For the second term, we are left with \cancel{(3)}(5)\cancel{(x)(x)(y)}(y)(y)=5y^2.
Therefore,
3x^5y+15x^2y^3=3x^2y(x^3+5y^2)
| NOTE |
| To get the GCF of each variable, we take the smallest exponent of that variable among all the terms. For instance, in this example, the variable x is raised to the power of 5 in the first term, and 2 in the second term. As 2 is smaller than 5, the GCF is x^2. For the variable y, the first term is raised to the power of 1 while the second term is raised to the power of 3. As 1 is smaller than 3, the GCF is y. |
Example 2
Factor out the GCF for the expression 7xy^2+63x^2y^3-49xy^5.
Solution
The expression 7xy^2+63x^2y^3-49xy^5 has three terms: 7xy^2, 63x^2y^3, and 49xy^5.
The GCF of 7, 63, and 49 is 7.
The GCF of x, x^2, and x is x.
The GCF of y^2, y^3, and y^5 is y^2.
Hence, the GCF of the three terms is 7xy^2.
After factoring out 7xy^2 from the first term, we are left with \cancel{(7)(x)(y)(y)}=1.
After factoring out 7xy^2 from the second term, we are left with \cancel{(7)(x)(y)(y)}(9)(x)(y)=9xy.
After factoring out 7xy^2 from the third term, we are left with \cancel{(7)(x)(y)(y)}(7)(y)(y)(y)=7y^3.
Therefore,
7xy^2+63x^2y^3-49xy^5=7xy^2(1+9xy-7y^3)
| NOTE |
| If all the factors of a term have been factored out, what remains is 1, not 0. For instance, in this example, after factoring out 7xy^2 from the first term, we are left with the factor 1. This is because 7xy^2 can be expressed as 7xy^2\times1, but not 7xy^2\times0. |
Example 3
Factor out the GCF for the expression 3x(x+2)+7(x+2).
Solution
For this example, there are two terms in the expression 3x(x+2)+7(x+2).
The first term is 3x(x+2) and the second is 7(x+2). We see that both terms have a common factor (x+2).
Therefore, we can factor this common factor out.
After factoring out (x+2) from the two terms, we are left with 3x\cancel{(x+2)}=3x and 7\cancel{(x+2)}=7, respectively.
Therefore,
3x(x+2)+7(x+2)=(x+2)(3x+7)
Example 4
Factor out the GCF for the expression x(5x-6)+(5x-6).
Solution
For this example, there are two terms in the given expression.
The first term is x(5x-6) and the second term is (5x-6). We see that both terms have a common factor (5x-6).
Therefore, we can factor this common factor out.
After factoring out (5x-6) from the two terms, we are left with x\cancel{(5x-6)}=x and 1\cancel{(5x-6)}=1, respectively.
Therefore,
x(5x-6)+(5x-6)=(5x-6)(x+1)
Once again, note that as the second term only has one factor and it has been factored out, what remains is 1, not 0.
Example 5
Factor out the GCF for the expression 7x(x-1)^3-3xy^2(x-1)^2.
Solution
Are you able to identify the two terms in the given expression? That’s right, the two terms are 7x(x-1)^3 and 3xy^2(x-1)^2.
The first term has the factors 7, x and (x-1)(x-1)(x-1). The second term has the factors 3, x, y^2 and (x-1)(x-1).
Therefore, the common factors are x and (x-1)^2.
After factoring x(x-1)^2 from the two terms, we are left with 7\cancel{x(x-1)(x-1)}(x-1)=7(x-1) and 3\cancel{x}y^2\cancel{(x-1)(x-1)}=3y^2, respectively.
Therefore,
\begin{aligned} 7x(x-1)^3-3xy^2(x-1)^2 &=x(x-1)^2[7(x-1)-3y^2]\\&=x(x-1)^2(7x-7-3y^2) \end{aligned}In the last step above, we expand the terms in the square brackets to simplify our answers.
Example 6
Factor out the GCF for the expression 9x(y-z)+7(z-y).
Solution
For this example, there are two terms in the given expression: 9x(y-z) and 7(z-y).
It seems like there aren’t any common factors among the two terms.
However, z-y is actually negative of y-z. In other words,
z-y=-(y-z)
(You can verify that the two sides are equal by expanding the right-hand side.)
Therefore,
\begin{aligned} 9x(y-z)+7(z-y)&=9x(y-z)+7[-(y-z)]\\&=9x(y-z)+[-7(y-z)]\\&= 9x(y-z)-7(y-z)\\&=(y-z)(9x-7)\end{aligned}Example 7
Factor out the GCF for the expression 3p(x-y)+8q(x+z).
Solution
This is a trick question as there is no common factor among the two terms in the given expression.
The two terms in the expression are 3p(x-y) and 8q(x+z). Although there is an x in both terms, we cannot factor it out as it is not a factor.
In other words,
\begin{aligned}3p(x-y)+8q(x+z)\ne x[3p(-y)+8q(z)] \end{aligned}Lesson 3: Difference of Squares
Like the name suggests, difference of squares is a technique for factoring a difference of two squares. This is done using the following equation:
a^2-b^2=(a+b)(a-b)
Try expanding the right-hand side of the equation to verify that the two sides are equal. Let’s look at some examples using the above formula to do factoring.
Example 1
Factor x^2-16 completely.
Solution
As 16=4^2, we have a difference of two squares (x^2 and 4^2) here. Therefore, a=x and b=4.
\begin{aligned}x^2-16&=x^2-4^2\\&=(x+4)(x-4) \end{aligned}Note:
To express 16 as a square, we can take its square root. \sqrt{16} = 4. Therefore, 16=4^2.
Example 2
Factor 3x^2-363 completely.
Solution
This does not seem to be a difference of squares question as 3x^2 and 363 are not perfect squares.
However, recall that we mentioned earlier that when doing factoring, we should always try to factor out the GCF first.
The GCF for 3x^2 and 363 is 3. Factoring out 3 from the two terms gives us two perfect squares.
The full solution is as follows:
\begin{aligned}3x^2-363&=3(x^2-121)\\&=3(x^2-11^2)\\&=3(x+11)(x-11) \end{aligned}Note:
\sqrt{121}=11. Therefore, 121=11^2.
Example 3
Factor (x-2)^2-(y+1)^2 completely.
Solution
Once again, we have a difference of two squares: (x-2)^2 and (y+1)^2.
a=(x-2) and b=(y+1)
The full solution is as follows:
\begin{aligned}&(x-2)^2-(y+1)^2\\=&[(x-2)+(y+1)][(x-2)-(y+1)]&&\text{(Note 1)}\\=&(x-2+y+1)(x-2-y-1)&&\text{(Note 2)}\\=&(x+y-1)(x-y-3) \end{aligned}
Note 1:
Factor the expression using the difference of squares formula. As a consists of 2 terms, we need to enclose it in a pair of parentheses (i.e., a=(x-2)); the same applies to b (b=(y+1)).
This means that we have to use square brackets to enclose a+b and a-b when we do our factoring.
Note 2:
Here, we simplify the expressions inside the square brackets after factoring. Check out this tutorial on simplifying expansions if you need help.
Example 4
Factor 8(2x-1)^2-18(x+4)^2 completely.
Solution
For this example, we need to express the two terms as perfect squares.
However, 8 and 18 are non-squares. Recall that we encountered something similar in Example 2? That’s right, we have to factor out the GCF first.
8 and 18 have a common factor of 2. Factoring out 2 from the two numbers give us 4 and 9, respectively, both of which are perfect squares.
Let’s consider 4(2x-1)^2; we can rewrite it as 2^2(2x-1)^2. In other words, a=2(2x-1).
For 9(x+4)^2, we can rewrite it as 3^2(x+4)^2. Therefore, b=3(x+4).
Hence,
\begin{aligned} &8(2x-1)^2-18(x+4)^2\\=&2[4(2x-1)^2-9(x+4)^2]&&\text{ (Note 1)}\\=&2[2(2x-1)+3(x+4)][2(2x-1)-3(x+4)]&&\text{ (Note 2)}\\=&2(4x-2+3x+12)(4x-2-3x-12)&&\text{ (Note 3)}\\=&2(7x+10)(x-14)\end{aligned}Notes:
- Factor out the GCF 2.
- Factor the expression using the difference of squares formula
- Simplify the expressions inside the square brackets after factoring.
Lesson 4: Cross Method
Cross method is a trial and error process for factoring expressions of the form ax^2+bx+c or ax^2+bxy+cy^2, where a, b and c are constants and x and y are variables.
The steps are as follows:
| Step | For ax^2+bx+c | For ax^2+bxy+cy^2 |
|---|---|---|
| 1 | Consider the terms ax^2 and c (i.e., the quadratic and constant terms). | Consider the terms ax^2 and cy^2 (i.e., the two quadratic terms). |
| 2 | Find pairs of factors of these two terms. Write the factors in two separate columns of a 2×2 grid and multiply the factors diagonally opposite each other. Add the products together. | |
| 3 | If the sum of the products is equal to the third term (i.e., bx), the grid is correct. The factors can then be obtained from the rows of the grid. | If the sum of the products is equal to the third term (i.e., bxy), the grid is correct. The factors can then be obtained from the rows of the grid. |
Let’s look at some examples.
Example 1
Factor the expression 12x^2-13x-14.
Solution
We need to consider factors of 12x^2 and -14.
First, we write the factors of 12x^2 and -14 in two separate columns. Next, we multiply the factors diagonally opposite each other and add the products together.
Let’s consider a few possible grids.
Grid 1
| 12x | -2 |
| x | 7 |
The first column lists two possible factors of 12x^2 (i.e., 12x and x).
The second column lists two possible factors of -14 (i.e., -2 and 7).
12x is diagonally across 7 and x is diagonally across -2.
- 12x\times 7=84x
- x\times(-2)=-2x
Adding the two products gives us 84x+(-2x)=82x.
As 82x is not equal to the remaining term (-13x), this grid is not the correct one.
Grid 2
| 3x | -2 |
| 4x | 7 |
The first column lists two possible factors of 12x^2 (i.e., 3x and 4x).
The second column lists two possible factors of -14 (i.e., -2 and 7).
3x is diagonally across 7 and 4x is diagonally across -2.
- 3x\times 7=21x
- 4x\times(-2)=-8x
Adding the two products gives us 21x+(-8x)=13x
As 13x is not equal to the remaining term (-13x), this grid is not the correct one.
However, the difference between 13x and -13x is just the sign. When we encounter such a situation, we can just switch the signs of both numbers in the second column to get the correct grid. Let’s do that now.
Grid 3
| 3x | 2 |
| 4x | -7 |
- 3x\times(-7)=-21x
- 4x\times 2=8x
Adding the two products gives us -21x+8x=-13x.
Great. This grid works as the sum of the two products equals the remaining term. Once we find a grid that works, the factors are given by the two rows in the grid.
In other words, the first factor is (3x+2) and the second factor is (4x-7).
Therefore,
12x^2-13x-14=(3x+2)(4x-7)
Example 2
Factor the expression -2x^2+15+x.
Solution
The two terms that we should analyze are -2x^2 and 15 (the quadratic and constant terms).
Let’s consider a few possible grids.
Grid 1
| -2x | 1 |
| x | 15 |
- -2x\times 15=-30x
- x\times1=x.
As -30x+x=-29x\neq x, this grid is not the correct one.
Grid 2
| -2x | 3 |
| x | 5 |
- -2x\times 5=-10x
- x\times3=3x.
As -10x+3x=-7x\neq x, this grid is not the correct one.
Grid 3
| -2x | 5 |
| x | 3 |
- -2x\times 3=-6x
- x\times5=5x.
As -6x+5x=-x\neq x, this grid is not the correct one.
However, the difference between x and -x is just the sign. Hence, we can get the correct grid by changing the signs of the two numbers in the second column. That gives us the grid below.
Grid 4
| -2x | -5 |
| x | -3 |
- -2x\times (-3)=6x
- x\times (-5)=-5x.
As 6x+(-5x)=x, this grid is correct. Therefore,
-2x^2+x+15=(-2x-5)(x-3)
Note that we can factor out the negative signs from the first factor and distribute it to the terms in the second factor.
Previously, we learned that distributing a negative factor toggles the signs of the terms that it is distributed to. The same happens when you factor out a negative factor or a negative sign.
Hence, we can also express the answer as
\begin{aligned} -2x^2+x+15&=(-2x-5)(x-3)\\&= -(2x+3)(x-3)\\&=(2x+3)[-(x-3)]\\&=(2x+3)(3-x)\end{aligned}Example 3
Factor the expression 14p^2-15q^2-11pq.
Solution
The two terms that we should analyze are 14p^2 and -15q^2 (the two quadratic terms).
Let’s consider a few possible grids.
Grid 1
| p | -q |
| 14p | 15q |
- p\times 15q=15pq
- 14p\times(-q)=-14pq.
As 15pq+(-14pq)=pq\neq -11pq, this grid is not the correct one.
Grid 2
| 7p | -5q |
| 2p | 3q |
- 7p\times 3q=21pq
- 2p\times(-5q)=-10pq.
As 21pq+(-10pq)=11pq\neq -11pq, this grid is not the correct one.
Grid 3
| 7p | 5q |
| 2p | -3q |
- 7p\times (-3q)=-21pq
- 2p\times5q=10pq.
As -21pq+10pq=-11pq, this grid is correct. Therefore,
14p^2-11pq-15q^2=(7p+5q)(2p-3q)
Lesson 5: Grouping
Grouping is a factoring technique that involves breaking down an expression into an addition of two or more simpler expressions. Let’s look at some examples.
Example 1
Factor completely pa-5pb+2aq-10bq.
Solution
The expression in this example has 4 terms. Let’s break it down into an addition of two simpler expressions. The best way to do this is to use parentheses to group the terms into two groups.
Let’s do that now.
pa-5bp+2aq-10bq=(pa-5bp)+(2aq-10bq)
After grouping the 4 terms into 2 groups, we can factor the groups separately.
The terms in the first group have a common factor p while those in the second group have a common factor 2q. Therefore,
\begin{aligned} pa-5bp+2aq-10bq &=(pa-5bp)+(2aq-10bq)\\&=p(a-5b)+2q(a-5b) \end{aligned}Now, we end up with two terms: p(a-5b) and 2q(a-5b).
(a-5b) is a common factor of both terms. Hence, we can factor that out further. Therefore, the final solution is
\begin{aligned} pa-5bp+2aq-10bq&=p(a-5b)+2q(a-5b)\\&=(a-5b)(p+2q) \end{aligned}Example 2
Factor completely 2px-2py-3qx+3qy.
Solution
Let’s do the grouping using parentheses first.
This expression consists of 4 terms, 2px, -2py, -3qx, \text{ and }3qy.
Therefore, we can group them into two groups as follows:
2px-2py-3qx+3qy=(2px-2py)+(-3qx+3qy)
We can factor out 2p from the first group.
From the second group, I’m going to factor out -3q instead of 3q. Recall that factoring out a negative factor changes the sign of the term.
Factoring out -3q from -3qx gives us \cancel{-3q}x=x.
Factoring out -3q from 3qy (which can be expressed as -(-3qy)) gives us -(\cancel{-3q}y)=-y.
Therefore,
\begin{aligned} 2px-2py-3qx+3qy&=(2px-2py)+(-3qx+3qy) \\&=2p(x-y)+[-3q(x-y)]\\&=2p(x-y)-3q(x-y)\\&=(x-y)(2p-3q) \end{aligned}Example 3
Factor completely 2ap-2aq+bq-bp.
Solution
\begin{aligned} 2ap-2aq+bq-bp&=(2ap-2aq)+(bq-bp)\\&=2a(p-q)+b(q-p) \end{aligned}Here, we end up with a situation very similar to Example 6 in Lesson 2. We know that q-p=-(p-q).
Therefore,
\begin{aligned} 2ap-2aq+bq-bp&=2a(p-q)+b(q-p)\\&=2a(p-q)+[-b(p-q)]\\&=2a(p-q)-b(p-q)\\&=(p-q)(2a-b) \end{aligned}Example 4
Factor completely 1-x^2-2xy-y^2.
Solution
In most cases, when we do grouping, we group the terms in the expression into equal groups. For instance, if there are 4 terms, we group them into 2 groups of 2 terms each. However, sometimes, there are exceptions. This example is one such exception.
If you try grouping the terms in this example into groups of 2, you will not be able to proceed further, as shown below:
\begin{aligned} 1-x^2-2xy-y^2&=(1-x^2)+(-2xy-y^2)\\&=(1+x)(1-x)+[-y(2x+y)]\\&=(1+x)(1-x)-y(2x+y)\end{aligned} After grouping, we end up with two groups: (1-x^2) and (-2xy-y^2).
The first group consists of a difference of squares. Hence, we can factor it using the difference of squares formula.
The second group consists of a common factor -y. Hence, we can factor it out.
However, after factoring the two groups, we end up with two terms that have no common factors. As a result, we are unable to proceed further.
The correct way to do this question is to group the last three terms together. The full solution is shown below:
\begin{aligned}1-x^2-2xy-y^2&=(1)+(-x^2-2xy-y^2)\\&=(1)-(x^2+2xy+y^2) &&\text{(Note 1)}\\&=(1)^2-(x+y)^2 &&\text{(Note 2)}\\&=[1+(x+y)][1-(x+y)]&&\text{(Note 3)}\\&=(1+x+y)(1-x-y) \end{aligned}Notes:
- Factor the negative signs out
- Factor (x^2+2xy+y^2) using the cross method
- We end up with a difference of squares in the previous step, where a=1, b=(x+y). Hence, we factor the expression using the formula a^2-b^2=(a+b)(a-b)