Factorization: Common factor extraction

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Factorization

The factorization we do by extracting the common factor is our way of modifying the way the exercise is written, that is, from an expression with addition to an expression with multiplication.

For example, the expression
2A+4B2A + 4B
is composed of two terms and a plus sign. We can factor it by excluding the largest common term.
In this case it is 2 2 .

We will write it as follows:
​​​​​​​2A+4B=2Γ—(A+2B)​​​​​​​2A + 4B = 2\times (A + 2B)

Since both terms ( A A and B B ) were multiplied by 2 2 we could "extract" it. The remaining expression is written in parentheses and the common factor (the 2 2 ) is kept out.
In this way we went from having two terms in an addition operation to having a multiplication. This procedure is called factorization.

You can also apply the distributive property to do a reverse process as needed.
In certain cases we will prefer to have a multiplication and in others an addition.

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Test yourself on factorization - common factor!

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Decompose the following expression into factors:

\( 36mn-60m \)

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In this article we will learn how to factor by extracting the common factor, that is, we will see how to go from an expression with addition to an expression with multiplication.

We will learn it through many examples with ascending levels of difficulty. We will learn how to extract a common factor that can be a number, unknown, expression in parentheses or other.

To solve exercises of this type you must handle very well the distributive property and the extended distributive property that will allow you to open expressions that are in parentheses. You must also know the law of exponents.

amn=amΓ—ana^{mn} = a^m \times a^n


What is the common factor?

In this article we will see how to go from an expression with several terms to one that includes only one.


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Example 1

Let's look at the expression:

2A+4B2A + 4B

This expression is composed of two terms. We can factor it by excluding the largest common term. In this case it is 22.
We will write it as follows:

2A+4B=2Γ—(A+2B)2A + 4B = 2\times (A + 2B)

We will notice that we had two addends and now we have a multiplication. This procedure is called factoring.
We can use the distributive property to do the process in reverse. We will multiply the 22 by each of the terms in parentheses:

2A+4B=2Γ—(A+2B)2A + 4B = 2\times (A + 2B)

In certain cases we will prefer an expression with multiplication and in others with addition.


Example 2

​​​​​​​4X+CX​​​​​​​4X + CX

The ​​​​​​​X​​​​​​​X is the greatest common factor between these two summands. We can write it like this:

XΓ—(4+C)X\times (4 + C)

Again, we obtained a multiplication.
Pay attention, to verify that we have factored correctly we will always look at the multiplication obtained and do the inverse procedure. If we get back to the original expression it means that we have done it right.
For example, in the last exercise we have obtained after factoring:

XΓ—(4+C)X\times (4 + C)

Let's make use of the distributive property and it will give us.

4X+CX4X + CX

This is the expression we started with, it means that we have done well.


Do you know what the answer is?

Example 3

Let's look at the expression

Z5+3Z7Z^5 + 3Z^7

To discover the common factor in this expression we must know the following formula well:

amn=amΓ—ana^{mn} = a^m \times a^n

Let's go back to our expression and we will see that we can write it like this:

Z5+3Z7=Z5+3Z5Γ—Z2Z^5 + 3Z^7 = Z^5 + 3Z^5\times Z^2

That is, we factor the expression 3Z23Z^2into 3Z5Γ—Z23Z^5\times Z^2 We did this since Z5Z^5 is the largest exponent that is common to both factors.
Now we can extract Z5Z^5 because it is the common factor and it will give us:

Z5+3Z7=Z5Γ—1+3Z5Γ—Z2=Z5Γ—(1+3Z2)Z^5+ 3Z^7 = Z^5\times 1 + 3Z^5\times Z^2 = Z^5\times (1 + 3Z^2)

We have obtained an expression with multiplication just as we wanted. Notice that, the expression Z5Z^5is equivalent to Z5Γ—1Z^5\times 1We chose to write it this way because it makes it easier for us to find the common factor. That is why the number 11 appears in parentheses.


Example 4: Common factor for more than two summands

In some cases we will come across an expression that has more than two summands, for example:

3A3+6A5+9A43A^3 + 6A^5 + 9A^4

The greatest common factor that we can extract from each of the terms is 3A33A^3 To see it in a clearer way we can write the exercise as follows:

3A3+6A5+9A4=3A3Γ—1+3A3Γ—2A2+3A3Γ—3A3A^3 + 6A^5 + 9A^4 = 3A^3 \times 1 + 3A^3 \times 2A^2 + 3A^3 \times 3A

After extracting the common factor 3A33A^3 the expression will look like this:

3A3Γ—(1+2A2+3A)3A^3 \times (1 + 2A^2 + 3A)

Here we have seen again, how we have started with an expression composed of several summands and have moved on to a multiplication.


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Example 5: Extraction of common factor for an expression in parentheses

Let's look at the following exercise:

3AΓ—(Bβˆ’5)+8Γ—(B–5)3A\times (B - 5) + 8 \times (B – 5)

We will notice that the expression (B–5)(B – 5) appears in both terms, therefore, we can extract it as a common factor.

After extracting the common factor it will give us:

3AΓ—(Bβˆ’5)+8Γ—(B–5)=(B–5)(3A+8)3A\times (B - 5) + 8\times (B – 5) = (B – 5) (3A + 8)


Example 6: Expressions with opposite signs in brackets

Let's look at the exercise:

3(Xβˆ’4)+X(4βˆ’X)3(X-4) + X(4-X)

At first glance it might confuse us and seem to us that there is no common factor that we can extract. But notice! The expressions

(Xβˆ’4) (X-4) and (4βˆ’X)(4-X) differ in the sign. That is, if we take one of them and multiply it by a βˆ’1-1 we will arrive at the other expression. Let's see it clearly with the distributive property:

βˆ’1Γ—(Xβˆ’4)=βˆ’X+4=4–X-1\times (X-4) = -X + 4 = 4 – X

Now let's go back to the original expression

3(Xβˆ’4)+X(4βˆ’X)3(X-4) + X(4-X)

We can write it as follows:

3(Xβˆ’4)+X(4βˆ’X)=3(Xβˆ’4)+XΓ—(βˆ’1)Γ—(Xβˆ’4)=3(Xβˆ’4)–X(Xβˆ’4)3(X-4) + X(4-X) = 3(X-4) + X\times (-1)\times (X-4) = 3(X-4) – X(X-4)

Now we can extract the common factor as in the previous exercises:

3(Xβˆ’4)+X(4βˆ’X)=3(Xβˆ’4)+XΓ—(βˆ’1)Γ—(Xβˆ’4)=(Xβˆ’4)(3–X)3(X-4) + X(4-X) = 3(X-4) + X\times (-1)\times (X-4) = (X-4)(3 – X)

Let's see what we have arrived at:

3(Xβˆ’4)+X(4βˆ’X)=(Xβˆ’4)(3–X)3(X-4) + X(4-X) = (X-4)(3 – X)

Remember! To verify the result you can do the reverse way, that is, take the last expression we obtained and get to the original one by means of the extended distributive property. Give it a try.


Do you think you will be able to solve it?

Example 7: Advanced level exercise

Factor the following expression:

3b2+3b+2b+23b^2 + 3b + 2b + 2

At first glance it would appear that there is no common factor among the four summands. Therefore, we will focus, separately, on the first two summands and then on the second two summands. We will write the expression as follows:

3b2+3b+2b+2=3bΓ—b+3bΓ—1+2Γ—b+2Γ—13b^2 + 3b + 2b + 2 = 3b\times b + 3b\times 1 + 2\times b + 2\times 1

Recall again that it is not necessary to write the multiplication by 11. At this stage we will only write it for our own convenience.
Now we will take out the common factor of the first two summands and, separately, we will take out the common factor of the second two summands, it will give us like this:

3b2+3b+2b+2=3bΓ—b+3bΓ—1+2Γ—b+2Γ—1=3b(b+1)+2(b+1)3b^2 + 3b + 2b + 2 = 3b\times b + 3b\times 1 + 2\times b + 2\times 1 = 3b (b + 1) + 2 (b + 1)

Let's see that now the expression (b+1)(b+1) appears twice, that means we can use it as a common factor.

We will take out a common factor one more time and we will obtain:

3b(b+1)+2(b+1)=(b+1)(3b+2)3b (b + 1) + 2 (b + 1) = (b+1)(3b+2)

In summary, we have obtained:

3b2+3b+2b+2=(b+1)(3b+2)3b2 + 3b + 2b + 2 = (b+1)(3b+2)

Let's notice that we have gone from an expression with four addends to a multiplication. Let's remember again that you can check your answers.
You can break down the last expression with the extended distributive property and check that you really get to the original expression.


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