# Factorization: Common factor extraction - Examples, Exercises and Solutions

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

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

Since both terms ( $A$ and $B$ ) were multiplied by $2$ we could "extract" it. The remaining expression is written in parentheses and the common factor (the $2$ ) 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.

## Examples with solutions for Factorization: Common factor extraction

### Exercise #1

Find the biggest common factor:

$12x+16y$

### Step-by-Step Solution

We begin by breaking down the coefficients 12 and 16 into multiplication exercises with a multiplying factor to eventually simplify:

$3\times4\times x+4\times4\times y$

We then extract 4 which is the common factor:

$4(3\times x+4\times y)=4(3x+4y)$

$4(3x+4y)$

### Exercise #2

Decompose the following expression into factors:

$20ab-4ac$

### Step-by-Step Solution

We first break down the coefficient of 20 into a multiplication exercise. That will help us to simplify the calculation :$5\times4\times a\times b-4\times a\times c$

We then extract 4a as a common factor:$4a(5\times b-c)=4a(5b-c)$

$4a(5b-c)$

### Exercise #3

Find the common factor:

$7a+14b$

### Step-by-Step Solution

We divide 14 into a multiplication exercise to help us simplify the calculation accordingly:$7\times a+7\times b\times2=$

We then extract the common factor 7:

$7(a+2\times b)=7(a+2b)$

$7(a+2b)$

### Exercise #4

Find the common factor:

$ab+bc$

### Step-by-Step Solution

$ab+bc=a\times b+b\times c$

If we consider that b is the common factor, it can be removed from the equation:

$b(ab+bc)=$

We divide by b:$b(\frac{ab}{b}+\frac{bc}{b})=$

$b(a+c)$

$b(a+c)$

### Exercise #5

Find the common factor:

$25y-100xy^2$

### Step-by-Step Solution

First, we will decompose the coefficients of the multiplication exercise that will help us find the common factor:

$25\times y-4\times25\times x\times y\times y$

Now find the common factor 25y:

$25y(1-4xy)$

$25y(1-4xy)$

### Exercise #6

Decompose the following expression into factors:

$4a+13b+58c$

### Step-by-Step Solution

Factor the given expression:

$4a+13b+58c$We will do this by extracting the greatest common factor, both from the numbers and the letters,

We will refer to the numbers and letters separately, remembering that a common factor is a factor (multiple) common to all terms of the expression,

We first notice that the numerical coefficients of the terms in the given expression, that is, the numbers 4, 13, 58, do not have a common factor, and this is because the number 13 is a prime number and the other two numbers are not multiples of it,

Therefore, there is no common factor for the numbers hence we select the number 1, as the common factor for the numbers (Reminder: a number raised to the power zero is always given to be equal to one)

For the letters:

There are three terms in the expression:
$a,\hspace{4pt}b,\hspace{4pt}c$It is easy to see that there is no common factor for these three terms,

Hence, it is not possible to factor the given expression with the help of a common factor.

Therefore, the correct answer is option d.

It is not possible to factorize the given expression by extracting the common factor.

### Exercise #7

Decompose the following expression into factors:

$15a^2+10a+5$

### Step-by-Step Solution

Factor the given expression:

$15a^2+10a+5$We will do this by factoring out the greatest common factor, both from the numbers and the letters,

We must refer to the numbers and letters separately, remembering that a common factor is a factor (multiplier) common to all the terms of the expression,

Note that the numerical coefficients of the terms in the given expression, that is, the numbers: 5,10,15 are all multiples of the number 5:

$15=3\cdot\underline{5}\\ 10=2\cdot\underline{5}\\$Therefore, the number 5 is the greatest common factor of the numbers,

For the letters:

Note that only the first two terms on the left depend on x, the third term is a free number that does not depend on x, hence there is no common factor for all three terms together for the letters (that is, we will consider the number 1 as the common factor for the letters)

Therefore, we can summarise as follows:

The greatest common factor (for numbers and letters together) is:

$5\cdot1\\ \downarrow\\ 5$Let's take the above value as a multiple outside the parenthesis and ask the question: "How many times must we multiply the common factor (including its sign) in order to obtain the terms of the original expression (including their signs)?" Using this method we can determine what is the expression inside the parenthesis that multiplied the common factor:

$\textcolor{red}{ 15a^2}\textcolor{blue}{+10a} \textcolor{green}{+5} \\ \underline{5}\cdot\textcolor{red}{3a^2}+\underline{5}\cdot\textcolor{blue}{(+2a)}+\underline{5}\cdot\textcolor{green}{(+1)}\\ \downarrow\\ \underline{5}(\textcolor{red}{3a^2}\textcolor{blue}{+2a}\textcolor{green}{+1})$In the previous expression, the operation is explained through colors and signs:

The common factor has been highlighted with an underscore, and the multiples inside the parenthesis are associated with the terms of the original expression with the help of colors.

Note that in the detail of the decomposition above, we refer both to the sign of the common factor (in black) that we extracted as a multiple outside of the parenthesis, as well as to the sign of the terms in the original expression (in colors) Note that there is no obligation to show it. Whilst the above method is described in stages it is recommended to jump directly to the broken down form in the last line whilst continuing to refer to the previous signs which are an integral part of the expression.

We can easily ensure that this decomposition is correct by opening the parentheses with the help of the distributive property. As such we can ensure that the original expression that we decomposed can be effectively recovered - Remember, this must be done emphasizing the sign of the members in the original expression as well as the sign (which is always selectable) of the common factor.

(Initially, you should use the previous colors to ensure you get all the terms and multiples in the original expression; later on, it is recommended not to use the colors)

Therefore, the correct answer is option b.

$5(3a^2+2a+1)$

### Exercise #8

Decompose the following expression into factors by removing the common factor:

$xyz+yzt+ztw+wtr$

### Step-by-Step Solution

Factor the given expression:

$xyz+yzt+ztw+wtr$
We will do this by extracting the highest common factor, both from the numbers and the letters.

We refer to the numbers and letters separately, remembering that a common factor is a factor (multiplier) common to all terms of the expression.

As the given expression does not have numeric coefficients (other than 1), we will look for the highest common factor of the letters:

There are four terms in the expression:
$xyz,\hspace{4pt}yzt,\hspace{4pt}ztw,\hspace{4pt}wtr$We will notice that in each of the four members there are three different letters, but there is not one or more letters that are included (in the multiplication) in all the terms; that is, there is no common factor for the four terms and therefore it is not possible to factor this expression by extracting a common factor.

Therefore, the correct answer is option d.

It is not possible to decompose the given expression into factors by extracting the common factor.

### Exercise #9

Fill in the missing value:

$?(18-2a)=10a-90$

### Step-by-Step Solution

Let's solve the given problem by factoring the expression on the right side and completing the missing part on the left side accordingly,

We will then examine each of the algebraic expressions in both sides of the given equation separately,

On the right side of the equation, the expression:

$10a-90$Let's now examine the expression on the left side of the equation:

$?(18-2a)$Let's note that in this expression there is a factor (unknown) multiplying an expression in parentheses, therefore to understand what this factor is - we'll return to the expression on the right side and factor it using common factor extraction,

In a routine manner - we'll look for first the largest common factor, we'll do this separately, for the numbers and for the letters:

In the expression on the right side, there are the numerical coefficients - 90 and 10, let's note that the number 90 is a multiple of the number 10:

$90=9\cdot10$Therefore, the number 10 is the largest common factor for the numbers,

Let's continue and examine the letters:

Let's note that only the left term in the expression on the right side, meaning the term -

$10a$depends on $a$unlike the second term in this expression:

$-90$which does not depend on $a$

Therefore, there is no common factor for these two terms (so we'll consider 1 as the common factor for the letters),

Let's summarize:

The largest common factor for numbers and letters together is:

$10\cdot1\\ \downarrow\\ 10$

Let's continue then and factor (using common factor extraction) the expression on the right side:

$\textcolor{red}{ 10a} \textcolor{blue}{-90 } \\ \underline{10}\cdot\textcolor{red}{a}+\underline{10}\cdot(\textcolor{blue}{ -9} )\\ \downarrow\\ \underline{10}\cdot(\textcolor{red}{a}\textcolor{blue}{ -9} )$

In the above expression, the operation is explained using colors and markings:

The common factor is highlighted with an underline, and the multipliers inside the parentheses are associated with the terms in the original expression using colors, let's note that we referred in the factoring details above both to the sign of the common factor (in black) that we extracted as a multiplier outside the parentheses and to the signs of the terms in the original expression (in colors), it's not necessary to present this in stages as described above, one can (and should) jump directly to the factored form in the last line, but we must definitely refer to the aforementioned signs, since in each term the sign is an inseparable part of it,

We can easily verify that this factoring is correct easily by opening the parentheses using the distributive property and verifying that indeed we get back the original expression we factored term by term, it's advisable to do this while emphasizing the signs of the terms in the original expression and the sign (always optional) of the common factor.

We factored the expression on the right side of the given equation, let's apply this factoring in the equation itself:

$?(18-2a)=10a-90 \\ \downarrow\\ ?(18-2a)=10(a-9)$

Let's note that the algebraic expressions in parentheses on both sides of the equality are not identical,

Therefore, we'll examine the difference between the expressions inside the parentheses, we want to bring the expression in parentheses on the right side to be identical to the expression in parentheses on the left side, let's note that between the two terms in parentheses on the left side and between the two terms in the expression in parentheses on the right side there is a clear proportion:

$\frac{18}{-9}=\frac{-2a}{a}=-2$ And additionally let's note that according to the multiplication rules:

$10=(-5)\cdot(-2)$Therefore, we can first rearrange (using the commutative property of addition) the expression in parentheses on the left side, in parallel we'll present the number 10 as a multiplication of numbers as mentioned above:

$10(a-9)\\ \downarrow\\ (-5)\cdot(-2)(-9+a)$Let's emphasize again that the sign preceding each term is an inseparable part of it, and therefore when using the commutative property (in addition) we switched the places of the terms along with their preceding signs.

Let's continue then and note that if we "insert" the factor $-2$into the parentheses, we'll get that the expressions in parentheses on both sides of the equality will be identical , we'll do this of course by applying the multiplication operation on the expression in parentheses, using the distributive property:

$x(y+z)=xy+xz$Let's apply this in the expression we got in the last stage, which is the expression on the right side in the given equation:

$(-5)\cdot(-2)(-9+a)\\ \downarrow\\ (-5)\big((-2)\cdot(-9)+(-2)\cdot a\big)\\ (-5)(18-2a)$In the last stage, we simplified the expression in parentheses,

Now let's return again to the original problem and summarize the development stages, we got that:

$?(18-2a)=10a-90 \\ \downarrow ?(18-2a)=10(a-9) \\ ?(18-2a)=(-5)(-2)(a-9) \\ ?(18-2a)=(-5)(18-2a) \\$

We got that the expressions in parentheses in the expressions on both sides of the equation are identical, therefore now we can easily complete the missing part (the factor marked on the left side with a question mark) and determine unequivocally that the correct answer is answer C.

Important note:

This problem is not an equation-solving problem, but a problem of completing the missing part, meaning - the solver is not required to find the value of the unknown (or unknowns) which when substituted in the equation will yield a true statement, but to find the missing algebraic expressions in the marked places and this in order to get equality between the algebraic expressions on both sides of the equation (i.e., regardless of the value of the unknown), therefore in the above problem (for example) although clearly for:

$a=9$indeed there is equality between the sides of the equation,

This information is of no use in order to answer what we were asked in the problem.

$-5$

### Exercise #10

Decompose the following expression into factors:

$14xyz+8x^2y^3z$

### Step-by-Step Solution

First, we break down all the powers into multiplication exercises and at the same time try to reduce the integers as much as possible:

7*2*xyz+2*4*x*x*y*y²*z

Now we use the substitution property to arrange the equation into a more manageable form:

2*x*y*z*7+2*x*y*z*x*y²

Lastly we try to find the common factor among all the parts - 2xyz

2xyz(7+xy²)

$2xyz(7+4xy^2)$

### Exercise #11

$2x^{90}-4x^{89}=0$

### Step-by-Step Solution

The equation in the problem is:

$2x^{90}-4x^{89}=0$Let's pay attention to the left side:

The expression can be broken down into factors by taking out a common factor, The greatest common factor for the numbers and letters in this case is $2x^{89}$since the power of 89 is the lowest power in the equation and therefore included both in the term where the power is 90 and in the term where the power is 89.

Any power higher than that is not included in the term where the power of 89 is the lowest, and therefore it is the term with the highest power that can be taken out of all the terms in the expression as a common factor for the variables.

For the numbers, note that the number 4 is a multiple of the number 2, so the number 2 is the greatest common factor for the numbers for the two terms in the expression.

Continuing and performing the factorization:

$2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0$Let's continue and remember that on the left side of the equation that was obtained in the last step there is an algebraic expression and on the right side the number is 0.

Since the only way to get the result 0 from a product is for at least one of the factors in the product on the left side to be equal to zero,

Meaning:

$2x^{89}=0 \hspace{8pt}\text{/}:2\\ x^{89}=0 \hspace{8pt}\text{/}\sqrt[89]{\hspace{6pt}}\\ \boxed{x=0}$

Or:

$x-2=0 \\ \boxed{x=2}$

In summary:

$2x^{90}-4x^{89}=0 \\ \downarrow\\ 2x^{89}(x-2)=0 \\ \downarrow\\ 2x^{89}=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$And therefore the correct answer is answer a.

$x=0,2$

### Exercise #12

Decompose the following expression into factors:

$(x+8)(2y+16)+4(3x+24)$

### Step-by-Step Solution

Breakdown into factors by groups The given expression:

$(x+8)(2y+16)+4(3x+24)$This is done by extracting a common factor, both for the numerators and for the letters, in one of the parts of the given expression (the parts of the expression are separated by addition or subtraction operations between the multiplication terms) or both, in order to be able to distinguish a common multi-variable factor, and take it out of the parentheses, etc.

Let's refer separately to the numerators and letters, recalling that a common factor is a factor (multiplier) common to all the terms in the expression,

Let's start by examining the two parts of the expression separately, the first expression on the left:

$(x+8)(2y+16)$and the second expression on the left:

$4(3x+24)$Note that the expression in the parentheses in the second expression above and the expression in the parentheses in the first expression on the left in the original expression are proportional to each other (i.e. it is possible to get from one expression to the other by multiplication by some factor), this is because it is possible to take out a common factor out of the parentheses:

$4(3x+24) \\ 4\cdot3(x+8) \\ 12(x+8)$

We used the fact that the number 24 is a multiple of the number 3:

$24=8\cdot3$

Let's now go back to the original expression in the question and apply this knowledge:

$(x+8)(2y+16)+4(3x+24) \\ \downarrow\\ (x+8)(2y+16)+4\cdot3(x+8)\\ (x+8)(2y+16)+12(x+8)$Let's now use the distributive property and rearrange the expression we got again:

$(x+8)(2y+16)+12(x+8) \\ (2y+16)(x+8)+12(x+8)$Now we can note that in the expression we got in the last step there is a common factor which is multi-variable (i.e. includes more than one variable in the expression):

$x+8$This is because it is a multiple of both the first part of the expression on the left (the parentheses multiplier) and of the second part of the expression on the left:

$(2y+16)\underline{(x+8)}+12\underline{(x+8)}$Therefore, we can take out this expression in its entirety out of the parentheses as a common factor and break down the given expression in the usual way (i.e. using the answer to the question: "By what can we multiply the common factor (including its sign) in order to get each of the terms in the original expression (including their sign)?"):

$\textcolor{red}{ (2y+16)(x+8)}\textcolor{blue}{+12(x+8) } \\ \underline{(x+8)}\cdot\textcolor{red}{(2y+16)}+\underline{(x+8)}\cdot\textcolor{blue}{(+12)}\\ \downarrow\\ \underline{(x+8)}\big(\textcolor{red}{(2y+16)}\textcolor{blue}{+12}\big)$

In the expression above the operation is explained using colors and signs:

The common factor is highlighted using an underline, and the multipliers inside the parentheses correspond to the terms in the original expression using colors, note that we also referred to the sign, both of the common factor (in black) that we took out of the parentheses and to the signs of the terms in the original expression (in colors), there is no need to display this in steps as described above, you can (and should) jump directly to the broken down form in the last line, but definitely need to refer to the signs above, as in each term the sign is an integral part of it,

So we got the expression broken down into factors (by groups):

$(x+8)\big((2y+16)+12\big)$Let's continue and complete the breakdown while simplifying the expression in the right parentheses, note that the expression we got in the right parentheses in the multiplier of the parentheses we got in the last step, can be further broken down into factors by taking out the common factor the number: 2, this is because the number 28 is a multiple of the number 2:

$(x+8)(2y+28) \\ (x+8)\cdot2(y+14) \\ 2(x+8)(y+14)$In the last step we used the distributive property to rearrange the expression we got.

Let's summarize the steps of breaking down the expression (by groups), we got that:

$(x+8)(2y+16)+4(3x+24) \\ (x+8)(2y+16)+12(x+8) \\ (x+8)\big((2y+16)+12\big) \\ (x+8)(2y+28) \\ 2(x+8)(y+14)$We can be sure that this breakdown is correct easily by opening the parentheses using the extended distribution law and verifying that indeed the original expression we broke down is obtained term by term, this should be done while paying attention to the signs of the terms in the original expression and to the sign (given for selection always) of the common factor.

Therefore, the correct answer is answer c.

$2(x+8)(y+14)$

### Exercise #13

Extract the common factor:

$4x^3+8x^4=$

### Step-by-Step Solution

First, we use the power law to multiply terms with identical bases:

$a^m\cdot a^n=a^{m+n}$It is necessary to keep in mind that:

$x^4=x^3\cdot x$Next, we return to the problem and extract the greatest common factor for the numbers separately and for the letters separately,

For the numbers, the greatest common factor is

$4$and for the letters it is:

$x^3$and therefore for the extraction

$4x^3$outside the parenthesis

We obtain the expression:

$4x^3+8x^4=4x^3(1+2x)$To determine what the expression inside the parentheses is, we use the power law, our knowledge of the multiplication table, and the answer to the question: "How many times do we multiply the common factor that we took out of the parenthesis to obtain each of the terms of the original expression that we factored?

Therefore, the correct answer is: a.

It is always recommended to review again and check that you get each and every one of the terms of the expression that is factored when opening the parentheses (through the distributive property), this can be done in the margin, on a piece of scrap paper, or by marking the factor we removed and each and every one of the terms inside the parenthesis, etc.

$4x^3(1+2x)$

### Exercise #14

Solve the following by removing a common factor:

$6x^6-9x^4=0$

### Step-by-Step Solution

First, we take out the smallest power

$6x^6-9x^4=$

$6x^4\left(x^2-1.5\right)=0$

If possible, we reduce the numbers by a common factor

Finally, we will compare the two sections with: $0$

$6x^4=0$

We divide by: $6x^3$

$x=0$

$x^2-1.5=0$

$x^2=1.5$

$x=\pm\sqrt{\frac{3}{2}}$



$x=0,x=\pm\sqrt{\frac{3}{2}}$

### Exercise #15

Decompose the following expression into factors:

$36mn-60m$

### Video Solution

$12m(3n-5)$