This may sound a bit dumb, but is there any significant difference between factoring, grouping trinomials, difference of squares, and GCF?

Nothing significant, but there are important (however small) differences and they are used for different things.

How would you work out the problem if there were only 3 terms?

Find two numbers that add to the middle coefficient and multiply to give the product of the first and last coefficients (or constants). This is called the ac method. Example: Factor 6x^2 + 19x + 10. 6*10 = 60, so we need to find two numbers that add to 19 and multiply to give 60. These numbers (after some trial and error) are 15 and 4. So split up 19x into 15x + 4x (or 4x + 15x), then factor by grouping: 6x^2 + 19x + 10 = 6x^2 + 15x + 4x + 10 = 3x(2x + 5) + 2(2x + 5) = (3x + 2)(2x + 5). Have a blessed, wonderful day!

In problem 3, I solved the expression and got the answer (4x+2)(2x+1.5). It said my answer was correct, but when I checked my answer, I got the same expression. What did I do wrong?

u used _1.5_ which is not a _natural_ number. Therefore answer is incorrect. Correct answer must include only natural numbers: (3x+2)(3x+4)

Hi Professor, Why dont you use the formula Ax^2+Bx-C where A*C=a*b AND B=a+b ? It helps me to learn it this way, since it opens up for future equations that take the square of A and C.

Hello! That's a great observation and method for understanding quadratic equations. The formula you mentioned, Ax^2 + Bx - C = 0, certainly has its benefits. It's a quadratic formula derived from completing the square. The quadratic formula you provided, where A*C = a*b and B = a + b, can be derived by equating the quadratic equation, ax^2 + bx + c = 0, with this particular form. By assuming that the equation has real roots, and then manipulating it further, you'll eventually arrive at the given formula. The advantage of using this form is that it allows you to see the connection between the quadratic equation and the coefficients A, B, and C. For example, you can notice the relationship between A and C and A^2 and C^2. This approach can be helpful for understanding various properties and relationships between the coefficients. In the end, both the traditional quadratic formula (-b ± √(b^2 - 4ac))/(2a) and the form you mentioned are mathematically equivalent. Using either form is a matter of personal preference and what you find most intuitive. It's great that you have found a method that helps you learn and understand quadratic equations better.

Lets say there is no GCF in anything. Then What?

There is always a GCF, it just depends on how many there are. Take 8 and 6, the factors of those two numbers are 1 and 2. 2 is bigger, so the GCF of 8 and 6 is 2. That works the same with every other set of numbers. Let's take 4 and 3. The factors of *3* are 1,3 The factors of *4* are 1,2,4 The only common factor here is 1, so 1 is the GCF of 4 and 3! So to answer your question, if you can't find the greatest common factor of two numbers, it's 1.

what the meaning of GCF?

GCF is the abbreviation for Greatest Common Factor. It is the value that you can evenly divide all terms by. It can be a number, a variable, or a mix of numbers and variables.

Why are they called quadratics? They are typically trinomials with a leading term raised to the second degree.

Quadratics actually are derived from the Latin word, 'quadratum', which literally means 'square'. A quadratic is basically a type of problem that deals with a variable multiplied by itself — an operation known as squaring. So, you could relate the word quadratic to Latin, not mathematics

Why does it say we cannot use grouping on this one 2x^2+3x+4x+12. It is usable though isn't it? Answer would be (x+2)(2x+3) right?

Your factors actually don't create the polynomial. If you multiply your factors, you will get: 2x^2+3x+4x+6, which does not match the original polynomial. Now, how do you know if isn't working before checking the factors by multiplying them. Pull the GCF from each pair of terms to get: 2x^2+3x+4x+12 = x(2x+3) + 4(x+3) Notice, the 2 binomials in parentheses do not match. To go to factors from this point, we need to remove a common binomial factor from each term. There is no common binomial factor. This tells use that the polynomial is not factorable. Hope this helps.

Main content

Course: Algebra 1 > Unit 13

Lesson 6: Factoring quadratics by grouping

Factoring by grouping

Google Classroom

Learn about a factorization method called "grouping." For example, we can use grouping to write 2x²+8x+3x+12 as (2x+3)(x+4).

What you need to know for this lesson

Factoring a polynomial involves writing it as a product of two or more polynomials. It reverses the process of polynomial multiplication.

We have seen several examples of factoring already. However, for this article, you should be especially familiar with taking common factors using the distributive property. For example,

6 x^{2} + 4 x = 2 x (3 x + 2)

What you will learn in this lesson

In this article, we will learn how to use a factoring method called grouping.

Example 1: Factoring $2 x^{2} + 8 x + 3 x + 12$ ‍

First, notice that there is no factor common to all terms in

2 x^{2} + 8 x + 3 x + 12

. However, if we group the first two terms together and the last two terms together, each group has its own GCF, or greatest common factor:

\underset{first grouping}{\underset{⏟}{(2 x^{2} + 8 x)}} + \underset{second grouping}{\underset{⏟}{(3 x + 12)}}

In particular, there is a GCF of

2 x

in the first grouping and a GCF of

3

in the second grouping. We can factor these out to obtain the following expression:

2 x (x + 4) + 3 (x + 4)

To factor

2 x

out of

2 x^{2} + 8 x

, we can divide each term by

2 x

The first term is: $\frac{2 x^{2}}{2 x} = x$ ‍
The second term is: $\frac{8 x}{2 x} = 4$ ‍

2 x^{2} + 8 x = 2 x (x + 4)

To factor

3

out of

3 x + 12

, we divide each term by

3

The first term is: $\frac{3 x}{3} = x$ ‍
The second term is: $\frac{12}{3} = 4$ ‍

3 x + 12 = 3 (x + 4)

Therefore, it follows that

2 x^{2} + 8 x + 3 x + 12 = 2 x (x + 4) + 3 (x + 4)

. A quick multiplication check can verify that the common factors were taken out correctly.

Notice that this reveals yet another common factor between the two terms:

x + 4

. We can use the distributive property to factor out this common factor.

2 x (x + 4) + 3 (x + 4) = (x + 4) (2 x + 3)

Since the polynomial is now expressed as a product of two binomials, it is in factored form. We can check our work by multiplying and comparing it to the original polynomial.

Example 2: Factoring $3 x^{2} + 6 x + 4 x + 8$ ‍

Let's summarize what was done above by factoring another polynomial.

\begin{aligned} 3 x^{2} + 6 x + 4 x + 8 \\ = (3 x^{2} + 6 x) + (4 x + 8) & Group terms \\ = 3 x (x + 2) + 4 (x + 2) & Factor out GCFs \\ = 3 x (x + 2) + 4 (x + 2) & Common factor! \\ = (x + 2) (3 x + 4) & Factor out x + 2 \end{aligned}

The factored form is

(x + 2) (3 x + 4)

Check your understanding

1) Factor $9 x^{2} + 6 x + 12 x + 8$ ‍.

2) Factor $5 x^{2} + 10 x + 2 x + 4$ ‍.

3) Factor $8 x^{2} + 6 x + 4 x + 3$ ‍.

Example 3: Factoring $3 x^{2} - 6 x - 4 x + 8$ ‍

Extra care should be taken when using the grouping method to factor a polynomial with negative coefficients.

For example, the steps below can be used to factor

3 x^{2} - 6 x - 4 x + 8

\begin{aligned} 3 x^{2} - 6 x - 4 x + 8 \\ (1) & = (3 x^{2} - 6 x) + (- 4 x + 8) & Group terms \\ (2) & = 3 x (x - 2) + (- 4) (x - 2) & Factor out GCFs \\ (3) & = 3 x (x - 2) - 4 (x - 2) & Simplify \\ (4) & = 3 x (x - 2) - 4 (x - 2) & Common factor! \\ (5) & = (x - 2) (3 x - 4) & Factor out x - 2 \end{aligned}

The factored form of the polynomial is

(x - 2) (3 x - 4)

. We can multiply the binomials to check our work.

A few of the steps above may seem different than what you saw in the first example, so you may have a few questions.

Where did the "+" sign between the groupings come from?

In step

(1)

, a "+" sign was added between the groupings

(3 x^{2} - 6 x)

and

(- 4 x + 8)

. This is because the third term

(- 4 x)

is negative, and the sign of the term must be included within the grouping.

Keeping the minus sign outside the second grouping is tricky. For example, a common error is to group

3 x^{2} - 6 x - 4 x + 8

(3 x^{2} - 6 x) - (4 x + 8)

. This grouping, however, simplifies to

3 x^{2} - 6 x - 4 x - 8

, which is not the same as the original expression.

Why factor out $- 4$ ‍ instead of $4$ ‍?

In step

(2)

, we factored out a

- 4

to reveal a common factor of

(x - 2)

between the terms. If we instead factored out a positive

4

, we would not obtain that common binomial factor seen above:

$\begin{aligned} (3 x^{2} - 6 x) + (- 4 x + 8) & = 3 x (x - 2) + 4 (- x + 2) \end{aligned}$ ‍

When the leading term in a group is negative, we will often need to factor out a negative common factor.

Check your understanding

4) Factor $2 x^{2} - 3 x - 4 x + 6$ ‍.

5) Factor $3 x^{2} + 3 x - 10 x - 10$ ‍.

6) Factor $3 x^{2} + 6 x - x - 2$ ‍.

Challenge problem

7*) Factor $2 x^{3} + 10 x^{2} + 3 x + 15$ ‍.

When can we use the grouping method?

The grouping method can be used to factor polynomials whenever a common factor exists between the groupings.

For example, we can use the grouping method to factor

3 x^{2} + 9 x + 2 x + 6

since it can be written as follows:

$\begin{aligned} (3 x^{2} + 9 x) + (2 x + 6) & = 3 x (x + 3) + 2 (x + 3) \end{aligned}$ ‍

We cannot, however, use the grouping method to factor

2 x^{2} + 3 x + 4 x + 12

because factoring out the GCF from both groupings does not yield a common factor!

$\begin{aligned} (2 x^{2} + 3 x) + (4 x + 12) & = x (2 x + 3) + 4 (x + 3) \end{aligned}$ ‍

Using grouping to factor trinomials

You can also use grouping to factor certain three termed quadratics (i.e. trinomials) like

2 x^{2} + 7 x + 3

. This is because we can rewrite the expression as follows:

$2 x^{2} + 7 x + 3 = 2 x^{2} + 1 x + 6 x + 3$ ‍

Then we can use grouping to factor

2 x^{2} + 1 x + 6 x + 3

(x + 3) (2 x + 1)

For more on factoring quadratic trinomials like these using the grouping method, check out our next article.

Want to join the conversation?

Sort by:

Nin
Posted 7 years ago. Direct link to Nin's post “This may sound a bit dumb...”
This may sound a bit dumb, but is there any significant difference between factoring, grouping trinomials, difference of squares, and GCF?
Button navigates to signup pageComment on Nin's post “This may sound a bit dumb...”
(40 votes)
Answer
- Copperhead514
  Posted 4 years ago. Direct link to Copperhead514's post “Nothing significant, but ...”
  Nothing significant, but there are important (however small) differences and they are used for different things.
  Button navigates to signup page
  (30 votes)
daleighellison
Posted 7 years ago. Direct link to daleighellison's post “How would you work out th...”
How would you work out the problem if there were only 3 terms?
Button navigates to signup pageComment on daleighellison's post “How would you work out th...”
(16 votes)
Answer
- Ian Pulizzotto
  Posted 7 years ago. Direct link to Ian Pulizzotto's post “Find two numbers that add...”
  Find two numbers that add to the middle coefficient and multiply to give the product of the first and last coefficients (or constants). This is called the ac method.
  
  Example: Factor 6x^2 + 19x + 10.
  6*10 = 60, so we need to find two numbers that add to 19 and multiply to give 60. These numbers (after some trial and error) are 15 and 4. So split up 19x into 15x + 4x (or 4x + 15x), then factor by grouping:
  
  6x^2 + 19x + 10 = 6x^2 + 15x + 4x + 10
  = 3x(2x + 5) + 2(2x + 5)
  = (3x + 2)(2x + 5).
  
  Have a blessed, wonderful day!
  Comment on Ian Pulizzotto's post “Find two numbers that add...”
  (76 votes)
rama
Posted 6 years ago. Direct link to rama's post “what the meaning of GCF?”
what the meaning of GCF?
Button navigates to signup pageComment on rama's post “what the meaning of GCF?”
(2 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “GCF is the abbreviation f...”
  GCF is the abbreviation for Greatest Common Factor.
  It is the value that you can evenly divide all terms by. It can be a number, a variable, or a mix of numbers and variables.
  Comment on Kim Seidel's post “GCF is the abbreviation f...”
  (24 votes)
louisaandgreta
Posted 3 years ago. Direct link to louisaandgreta's post “Why are they called quadr...”
Why are they called quadratics? They are typically trinomials with a leading term raised to the second degree.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- lemonomadic
  Posted 3 years ago. Direct link to lemonomadic's post “Quadratics actually are d...”
  Quadratics actually are derived from the Latin word, 'quadratum', which literally means 'square'.
  A quadratic is basically a type of problem that deals with a variable multiplied by itself — an operation known as squaring.
  So, you could relate the word quadratic to Latin, not mathematics
  Comment on lemonomadic's post “Quadratics actually are d...”
  (18 votes)
Seraj shaheen
Posted 10 months ago. Direct link to Seraj shaheen's post “Wow I got this better tha...”
Wow I got this better than how I learned it in school thanks Sal!
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
Dannon Foster
Posted a year ago. Direct link to Dannon Foster's post “Lets say there is no GCF ...”
Lets say there is no GCF in anything. Then What?
Button navigates to signup pageComment on Dannon Foster's post “Lets say there is no GCF ...”
(3 votes)
Answer
- AuroraAlberts
  Posted a year ago. Direct link to AuroraAlberts's post “There is always a GCF, it...”
  There is always a GCF, it just depends on how many there are.
  Take 8 and 6, the factors of those two numbers are 1 and 2.
  2 is bigger, so the GCF of 8 and 6 is 2.
  That works the same with every other set of numbers.
  
  Let's take 4 and 3.
  The factors of 3 are 1,3
  The factors of 4 are 1,2,4
  
  The only common factor here is 1, so 1 is the GCF of 4 and 3!
  
  So to answer your question, if you can't find the greatest common factor of two numbers, it's 1.
  Button navigates to signup page
  (8 votes)
Hafsa Mohamed
Posted 6 years ago. Direct link to Hafsa Mohamed's post “In problem 3, I solved th...”
In problem 3, I solved the expression and got the answer (4x+2)(2x+1.5). It said my answer was correct, but when I checked my answer, I got the same expression. What did I do wrong?
Button navigates to signup pageComment on Hafsa Mohamed's post “In problem 3, I solved th...”
(6 votes)
Answer
- smitters
  Posted a year ago. Direct link to smitters's post “u used _1.5_ which is not...”
  u used 1.5 which is not a natural number. Therefore answer is incorrect. Correct answer must include only natural numbers: (3x+2)(3x+4)
  Button navigates to signup page
  (2 votes)
Altangerel Chinzorig
Posted 2 years ago. Direct link to Altangerel Chinzorig's post “Why does it say we cannot...”
Why does it say we cannot use grouping on this one 2x^2+3x+4x+12. It is usable though isn't it? Answer would be (x+2)(2x+3) right?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 2 years ago. Direct link to Kim Seidel's post “Your factors actually don...”
  Your factors actually don't create the polynomial. If you multiply your factors, you will get: 2x^2+3x+4x+6, which does not match the original polynomial.
  
  Now, how do you know if isn't working before checking the factors by multiplying them.
  
  Pull the GCF from each pair of terms to get:
  2x^2+3x+4x+12 = x(2x+3) + 4(x+3)
  
  Notice, the 2 binomials in parentheses do not match. To go to factors from this point, we need to remove a common binomial factor from each term. There is no common binomial factor. This tells use that the polynomial is not factorable.
  
  Hope this helps.
  Comment on Kim Seidel's post “Your factors actually don...”
  (11 votes)
tarasandbeck
Posted 7 months ago. Direct link to tarasandbeck's post “Hi Professor, Why dont yo...”
Hi Professor,
Why dont you use the formula Ax^2+Bx-C where A*C=a*b AND B=a+b ? It helps me to learn it this way, since it opens up for future equations that take the square of A and C.
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Gonzalo Sandmeier
  Posted 7 months ago. Direct link to Gonzalo Sandmeier's post “Hello! That's a great obs...”
  Hello! That's a great observation and method for understanding quadratic equations. The formula you mentioned, Ax^2 + Bx - C = 0, certainly has its benefits. It's a quadratic formula derived from completing the square.
  
  The quadratic formula you provided, where A*C = a*b and B = a + b, can be derived by equating the quadratic equation, ax^2 + bx + c = 0, with this particular form. By assuming that the equation has real roots, and then manipulating it further, you'll eventually arrive at the given formula.
  
  The advantage of using this form is that it allows you to see the connection between the quadratic equation and the coefficients A, B, and C. For example, you can notice the relationship between A and C and A^2 and C^2. This approach can be helpful for understanding various properties and relationships between the coefficients.
  
  In the end, both the traditional quadratic formula (-b ± √(b^2 - 4ac))/(2a) and the form you mentioned are mathematically equivalent. Using either form is a matter of personal preference and what you find most intuitive. It's great that you have found a method that helps you learn and understand quadratic equations better.
  Button navigates to signup page
  (2 votes)
Joshua Zanetta
Posted a year ago. Direct link to Joshua Zanetta's post “I'm getting this!”
I'm getting this!
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- YourLordJoyBoy89
  Posted 16 days ago. Direct link to YourLordJoyBoy89's post “I want to get it too, how...”
  I want to get it too, how can I though?
  Button navigates to signup page
  (1 vote)

Algebra 1

Course: Algebra 1 > Unit 13

What you need to know for this lesson

What you will learn in this lesson

Example 1: Factoring 2x2+8x+3x+12‍

Example 2: Factoring 3x2+6x+4x+8‍

Check your understanding

Example 3: Factoring 3x2−6x−4x+8‍

Check your understanding

Challenge problem

When can we use the grouping method?

Using grouping to factor trinomials

Want to join the conversation?

Example 1: Factoring $2 x^{2} + 8 x + 3 x + 12$ ‍

Example 2: Factoring $3 x^{2} + 6 x + 4 x + 8$ ‍

Example 3: Factoring $3 x^{2} - 6 x - 4 x + 8$ ‍