Extracting the common factor in parentheses

šŸ†Practice common factoring

Common Factor Extraction Method:
Identify the largest free number that we can extract.
Then, let's move on to the variables and ask what is the least number of times the X X appears in any element?
Multiply the free number by the variable the same number of times we have found and we will obtain the greatest common factor.

To verify that you have correctly extracted the common factor, open the parentheses and see if you have returned to the original exercise.

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Test yourself on common factoring!

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\( x^4+2x^2=0 \)

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Factoring out the greatest common factor is the first operation we try to carry out when we want to break down an expression into factors.
We can remove from the parentheses a factor that is common to both elements and leave inside a simple and comfortable expression.
The greatest common factor is the largest factor that is completely common to both elements.


Operation steps for extracting the common factor

Notice which is the largest free number we can extract.
Then, let's move on to the unknowns and ask what is the least number of times that X X appears in any element?
Multiply the free number by the unknown the number of times we have found and we will obtain the greatest common factor. To verify that you have extracted the common factor correctly, open the parentheses and see if you have arrived at the original exercise.

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Let's look at an example of factoring out the common factor from the parentheses.

8x2+4x=8x^2+4x=
Let's see what is the greatest common factor we can take out, the answer is 44.
Now let's move on to the variable xx. What is the minimum number of times that xx appears in any term? The answer is 11.
Now let's multiply the common factor obtained by the variable the number of times we have found and it will give us the greatest common factor.
That is: 4Ɨx=4x4\times x=4x
Let's take out 4x4x as the common factor and we will obtain:
4x(2x+1)4x(2x+1)
This is our factorization.
When we want to find the solutions we will compare it with00 and it will give us:
4x(2x+1)=04x(2x+1)=0
X=0X=0
​​​​​​​2x+1=0​​​​​​​2x+1=0
2x=āˆ’12x=-1
x=āˆ’1/2x=-1/2
Therefore:
x=0,āˆ’1/2x=0,-1/2

Don't worry, as you practice extracting the common factor you will not need to act according to the operation steps as we taught them, you will do the extraction of the common factor intuitively and quickly.


Examples and exercises with solutions on factoring by taking out the common factor from the parentheses

Exercise #1

x4+2x2=0 x^4+2x^2=0

Video Solution

Step-by-Step Solution

To solve the equation x4+2x2=0x^4 + 2x^2 = 0, we will use the technique of factoring. Let's proceed step-by-step:

First, notice that both terms x4x^4 and 2x22x^2 have a common factor of x2x^2. We can factor x2x^2 out from the equation:

x2(x2+2)=0x^2(x^2 + 2) = 0

Now, to solve for xx, we apply the Zero Product Property, which gives us that at least one of the factors must be zero:

  • x2=0x^2 = 0 or
  • x2+2=0x^2 + 2 = 0

Solving the first case, x2=0x^2 = 0:

x=0x = 0

For the second case, x2+2=0x^2 + 2 = 0, we reach:

x2=āˆ’2x^2 = -2

Since x2=āˆ’2x^2 = -2 has no real solutions (squares of real numbers are non-negative), we can conclude that this equation doesn't provide additional real solutions.

Therefore, the only real solution to the given equation is x=0x = 0.

The correct choice from the provided options is:

x=0 x=0

Answer

x=0 x=0

Exercise #2

Solve the following problem:

x2āˆ’x=0 x^2-x=0

Video Solution

Step-by-Step Solution

Shown below is the given equation:

x2āˆ’x=0 x^2-x=0

First note that on the left side we are able to factor the expression using a common factor. The largest common factor for the numbers and letters in this case is x x and this is due to the fact that the first power is the lowest power in the equation. Therefore it is included both in the term with the second power and in the term with the first power. Any power higher than this is not included in the term with the first power, which is the lowest. Hence this is the term with the highest power that can be factored out as a common factor from all terms in the expression. Continue to factor the expression:

x2āˆ’x=0↓x(xāˆ’1)=0 x^2-x=0 \\ \downarrow\\ x(x-1)=0

Proceed to the left side of the equation that we obtained in the last step. There is a multiplication of algebraic expressions and on the right side the number 0. Therefore given that the only way to obtain 0 from a multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

x=0 \boxed{x=0}

or:

xāˆ’1=0↓x=1 x-1=0\\ \downarrow\\ \boxed{x=1}

Let's summarize then the solution to the equation:

x2āˆ’x=0↓x(xāˆ’1)=0↓x=0→x=0xāˆ’1=0→x=1↓x=0,1 x^2-x=0 \\ \downarrow\\ x(x-1)=0 \\ \downarrow\\ x=0 \rightarrow\boxed{ x=0}\\ x-1=0 \rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}

Therefore the correct answer is answer B.

Answer

x=0,1 x=0,1

Exercise #3

Solve the following problem:

7x3āˆ’x2=0 7x^3-x^2=0

Video Solution

Step-by-Step Solution

Solve the given equation:

7x3āˆ’x2=0 7x^3-x^2=0

Note that on the left side we are able to factor the expression using a common factor. The largest common factor for the numbers and letters in this case is x2 x^2 since the second power is the lowest power in the equation and therefore is included both in the term with the third power and in the term with the second power. Any power higher than this is not included in the term with the second power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression.

7x3āˆ’x2=0↓x2(7xāˆ’1)=0 7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0

Note that the left side of the equation that we obtained in the last step is a multiplication of algebraic expressions and on the right side the number 0.

Therefore, given that the only way to obtain 0 from a multiplication operation is to multiply by 0. Hence at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

x2=0/x=±0x=0 x^2=0 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0} (in this case taking the even root of the right side of the equation will indeed yield two possibilities, positive and negative. However since we're dealing with zero, we'll get only one possibility)

or:

7xāˆ’1=0 7x-1=0 Let's solve this equation in order to obtain the additional solutions (if they exist) to the given equation:

We obtained a simple first-degree equation which we'll solve by isolating the unknown on one side, we'll do this by moving terms and then dividing both sides of the equation by the coefficient of the unknown:

7xāˆ’1=07x=1/:7x=17 7x-1=0 \\ 7x=1\hspace{8pt}\text{/}:7\\ \boxed{x=\frac{1}{7}}

Let's summarize the solution of the equation:

7x3āˆ’x2=0↓x2(7xāˆ’1)=0↓x2=0→x=07xāˆ’1=0→x=17↓x=0,17 7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0\\ \downarrow\\ x^2=0 \rightarrow\boxed{ x=0}\\ 7x-1=0\rightarrow \boxed{x=\frac{1}{7}}\\ \downarrow\\ \boxed{x=0,\frac{1}{7}}

Therefore the correct answer is answer C.

Answer

x=0,x=17 x=0,x=\frac{1}{7}

Exercise #4

Solve the following equation:

7x10āˆ’14x9=0 7x^{10}-14x^9=0

Video Solution

Step-by-Step Solution

Shown below is the given equation:

7x10āˆ’14x9=0 7x^{10}-14x^9=0

First, note that on the left side we are able to factor the expression using a common factor.

The largest common factor for the numbers and variables in this case is 7x9 7x^9 given that the ninth power is the lowest power in the equation and therefore is included in both the term with the tenth power and the term with the ninth power. Any power higher than this is not included in the term with the ninth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 14 is a multiple of 7, therefore 7 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

7x10āˆ’14x9=0↓7x9(xāˆ’2)=0 7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0

On the left side of the equation that we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to obtain a result of 0 from a multiplication operation is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

7x9=0/:7x9=0/9x=0 7x^9=0 \hspace{8pt}\text{/}:7\\ x^9=0 \hspace{8pt}\text{/}\sqrt[9]{\hspace{6pt}}\\ \boxed{x=0}

In solving the equation above, we first divided both sides of the equation by the term with the variable, and then we proceeded to extract a ninth root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

xāˆ’2=0x=2 x-2=0 \\ \boxed{x=2}

Let's summarize the solution of the equation:

7x10āˆ’14x9=0↓7x9(xāˆ’2)=0↓7x9=0→x=0xāˆ’2=0→x=2↓x=0,2 7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0\\ \downarrow\\ 7x^9=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}

Therefore, the correct answer is answer A.

Answer

x=2,x=0 x=2,x=0

Exercise #5

x4+x2=0 x^4+x^2=0

Video Solution

Step-by-Step Solution

The problem at hand is to solve the equation x4+x2=0 x^4 + x^2 = 0 .

Let's begin by factoring the expression:

The given equation is: x4+x2=0 x^4 + x^2 = 0

We can factor out the common factor of x2 x^2 from both terms:

x2(x2+1)=0 x^2(x^2 + 1) = 0

To solve for x x , we set each factor equal to zero:

  • x2=0 x^2 = 0

Solving for x x , we have:

x=0 x = 0

Next, consider the second factor:

  • x2+1=0 x^2 + 1 = 0

Solving for x x , we have:

x2=āˆ’1 x^2 = -1

Since x2=āˆ’1 x^2 = -1 has no real solutions, we ignore these solutions in the real number system.

Thus, the only real solution to the equation x4+x2=0 x^4 + x^2 = 0 is:

x=0 x = 0

Answer

x=0 x=0

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