So, this metod is not applicable to this equation: 3x^2+5x+10 for example. What method can I use to factorize this equation? Or there are any quadratics can´t be factorizated?

The quadratic formula always works, so first look at the discriminant b^2 - 4 a c For your equation 5*2 - 4(3)(10) or 25 - 120 = - 95 Since you cannot take SQRT of negative number in real domain This cannot be factorable since there is no real solution

On question *4*)factor *3x^2 -2x- 5* In working it out I got the answer (x-1)(3x+5) where as the correct answer was (3x-5)(x+1). The difference appears because there were multiple ways to split the GCF. As far as I can tell they both give the right answer and expand back into 3x^2-2x-5. Is this typical when solving these problems or did I do something wrong or is this just a random special case that happens to work?

Sorry, but your version creates the wrong sign on the middle term. You can see this if you multiply your factors. You get: 3x^2+5x-3x-5. Notice: 5x-3x = 2x, not -2x. Hope this helps.

How would I solve a question like this: (2(1-x))/3 - 8/1=1/(6x)-(2x-3)/3

Multiply both sides by 3, which gets -2x-22=(1/2x)-2x-3 Add 2x, so -22=(1/2x)-3. Then add 3, so -19=(1/2x). Multiply by 2x so -38x=1. Then divide by -38, so x = (-1 / 38)

What if the resulting grouping ends up with contradicting signs inside the parenthesis (like 2x(x-4) -5(1+4) ) This is just an example not the actual question I encountered.

I'm assuming you mean 2x(x-4)-5(x+4) You just change the sign of the number you are factoring out. For example, in the second part, you have -5(x+4). Multiplying it you get -5x-20. To get (x-4) you would factor out 5 instead of -5, to get 5(x-4) Now you have 2x(x-4)+5(x-4) which then becomes (2x+5)(x-4) You can also do this to the first part, 2x(x-4) switch 2x to -2x then you get -2x(x+4)-5(x+4) then (-2x-5)(x+4).

What is the easiest way how to remember how to factor out quadratics?

There are some songs on youtube to help memorize these formulas. I recommend the quadratic formula set to the tune of "Pop Goes the Weasel", because you can use that formula for a lot of different equations, and the song is very memorable. Hope this answers your question :)

the letters are really confusing me im getting mixed up on ac and bc and what this and that equals can someone help me pls

Breath, focus, and concentrate. You can do this! Try writing it down and giving the letters different colours.

A question about the first problem. My answer is "(x+2)(3x+4)", but there are no choices in there. Even if the values of a or b are opposite, is the answer the same?

Your answer is correct (you can always check it by multiplying the 2 binomials). It matches option B in the answer list. Remember, the commutative property of multiplication tells us that 2(3) = 3(2). This property extends to any multiplication problem, including your binomials. (3x+4)(x+2) is the same as (x+2)(3x+4). Hope this helps.

Main content

Course: Algebra 1 > Unit 13

Lesson 6: Factoring quadratics by grouping

Factoring quadratics: leading coefficient ≠ 1

Google Classroom

Learn how to factor quadratic expressions as the product of two linear binomials. For example, 2x²+7x+3=(2x+1)(x+3).

What you need to know before taking this lesson

The grouping method can be used to factor polynomials with

4

terms by taking out common factors multiple times. If this is new to you, you'll want to check out our Intro to factoring by grouping article.

We also recommend that you review our article on factoring quadratics with a leading coefficient of 1 before proceeding.

What you will learn in this lesson

In this article, we will use grouping to factor quadratics with a leading coefficient other than

1

, like

2 x^{2} + 7 x + 3

Example 1: Factoring $2 x^{2} + 7 x + 3$ ‍

Since the leading coefficient of

(2 x^{2} + 7 x + 3)

2

, we cannot use the sum-product method to factor the quadratic expression.

Instead, to factor

2 x^{2} + 7 x + 3

, we need to find two integers with a product of

2 \cdot 3 = 6

(the leading coefficient times the constant term) and a sum of

7

(the

x

-coefficient).

Since

1 \cdot 6 = 6

and

1 + 6 = 7

, the two numbers are

1

and

6

To find these numbers, we can create a list of all numbers that multiply to

6

. Then, within that list, we look for those that sum to

7

Product: $m \cdot n = 6$ ‍	‍	Sum: $m + n = 7$ ‍
$1 \cdot 6 = 6$ ‍		$1 + 6 = 7$ ‍
$2 \cdot 3 = 6$ ‍		$2 + 3 = 5$ ‍
$(- 1) \cdot (- 6) = 6$ ‍		$(- 1) + (- 6) = - 7$ ‍
$(- 2) \cdot (- 3) = 6$ ‍		$\begin{aligned} (- 2) + (- 3) = - 5 \end{aligned}$ ‍

These two numbers tell us how to break up the

x

-term in the original expression. So we can express our polynomial as

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

We can now use grouping to factor the polynomial:

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

The factored form is

(2 x + 1) (x + 3)

Suppose we wrote the polynomial as

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

and then grouped as follows:

$\begin{aligned} 2 x^{2} + 6 x + 1 x + 3 \\ = (2 x^{2} + 6 x) + (1 x + 3) & Group terms \end{aligned}$ ‍

Notice that this grouping would also lead to the correct factorization.

$\begin{aligned} = 2 x (x + 3) + 1 (x + 3) & Factor out GCFs \\ = 2 x (x + 3) + 1 (x + 3) & Common factor! \\ = (x + 3) (2 x + 1) & Factor out x + 3 \\ = (2 x + 1) (x + 3) & Mult. is commutative \end{aligned}$ ‍

As long as you find the correct way to break up the

x

-term, it doesn't matter which piece you put with the first grouping and which you put with the second grouping.

We can check our work by showing that the factors multiply back to

2 x^{2} + 7 x + 3

Summary

In general, we can use the following steps to factor a quadratic of the form

a x^{2} + b x + c

Start by finding two numbers that multiply to $a c$ ‍ and add to $b$ ‍.
Use these numbers to split up the $x$ ‍-term.
Use grouping to factor the quadratic expression.

Check your understanding

1) Factor $3 x^{2} + 10 x + 8$ ‍.

To factor

3 x^{2} + 10 x + 8

, we need to find two integers with a product of

3 \cdot 8 = 24

and a sum of

10

To find these numbers, we can create a list of all numbers that multiply to

24

. Then, within that list, we look for those that sum to

10

Product: $m \cdot n = 24$ ‍	‍	Sum: $m + n = 10$ ‍
$1 \cdot 24 = 24$ ‍		$1 + 24 = 25$ ‍
$2 \cdot 12 = 24$ ‍		$2 + 12 = 14$ ‍
$3 \cdot 8 = 24$ ‍		$3 + 8 = 11$ ‍
$4 \cdot 6 = 24$ ‍		$4 + 6 = 10$ ‍

(Note that we do not need to list the negative factors here, since they will give us a negative sum!)

We can now break the term

10 x

into

4 x + 6 x

and use grouping to factor the polynomial:

\begin{aligned} 3 x^{2} + 10 x + 8 \\ = 3 x^{2} + 4 x + 6 x + 8 \\ = (3 x^{2} + 4 x) + (6 x + 8) & Group terms \\ = x (3 x + 4) + 2 (3 x + 4) & Factor out GCFs \\ = x (3 x + 4) + 2 (3 x + 4) & Common factor! \\ = (3 x + 4) (x + 2) & Factor out 3 x + 4 \end{aligned}

The factored form is

(3 x + 4) (x + 2)

2) Factor $4 x^{2} + 16 x + 15$ ‍.

To factor

4 x^{2} + 16 x + 15

, we need to find two integers with a product of

4 \cdot 15 = 60

and a sum of

16

To find these numbers, we can create a list of all numbers that multiply to

60

. Then, within that list, we look for those that sum to

16

Product: $m \cdot n = 60$ ‍	‍	Sum: $m + n = 16$ ‍
$1 \cdot 60 = 60$ ‍		$1 + 60 = 61$ ‍
$2 \cdot 30 = 60$ ‍		$2 + 30 = 32$ ‍
$4 \cdot 15 = 60$ ‍		$4 + 15 = 19$ ‍
$5 \cdot 12 = 60$ ‍		$5 + 12 = 17$ ‍
$6 \cdot 10 = 60$ ‍		$6 + 10 = 16$ ‍

(Note that we do not need to list the negative factors here, since they will give us a negative sum!)

We can now break the term

16 x

into

6 x + 10 x

and use grouping to factor the polynomial:

\begin{aligned} 4 x^{2} + 16 x + 15 \\ = 4 x^{2} + 6 x + 10 x + 15 \\ = (4 x^{2} + 6 x) + (10 x + 15) & Group terms \\ = 2 x (2 x + 3) + 5 (2 x + 3) & Factor out GCFs \\ = 2 x (2 x + 3) + 5 (2 x + 3) & Common factor! \\ = (2 x + 3) (2 x + 5) & Factor out 2 x + 3 \end{aligned}

The factored form is

(2 x + 3) (2 x + 5)

Example 2: Factoring $6 x^{2} - 5 x - 4$ ‍

To factor

6 x^{2} - 5 x - 4

, we need to find two integers with a product of

6 \cdot (- 4) = - 24

and a sum of

- 5

Since

3 \cdot (- 8) = - 24

and

3 + (- 8) = - 5

, the numbers are

3

and

- 8

To find these numbers, we can create a list of all numbers that multiply to

- 24

. Then, within that list, we look for those that sum to

- 5

Product: $m \cdot n = - 24$ ‍	‍	Sum: $m + n = - 5$ ‍
$1 \cdot (- 24) = - 24$ ‍		$1 + (- 24) = - 23$ ‍
$- 1 \cdot 24 = - 24$ ‍		$- 1 + 24 = 23$ ‍
$2 \cdot (- 12) = - 24$ ‍		$2 + (- 12) = - 10$ ‍
$- 2 \cdot 12 = - 24$ ‍		$- 2 + 12 = 10$ ‍
$3 \cdot (- 8) = - 24$ ‍		$3 + (- 8) = - 5$ ‍
$- 3 \cdot 8 = - 24$ ‍		$- 3 + 8 = 5$ ‍
$4 \cdot (- 6) = - 24$ ‍		$4 + (- 6) = - 2$ ‍
$- 4 \cdot 6 = - 24$ ‍		$\begin{aligned} - 4 + 6 = 2 \end{aligned}$ ‍

We can now write the term

- 5 x

as the sum of

3 x

and

- 8 x

and use grouping to factor the polynomial:

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

The factored form is

(2 x + 1) (3 x - 4)

We can check our work by showing that the factors multiply back to

6 x^{2} - 5 x - 4

Take note: In step

(1)

above, notice that because the third term is negative, a "+" was inserted between the groupings to keep the expression equivalent to the original. Also, in step

(2)

, we needed to factor out a negative GCF from the second grouping to reveal a common factor of

2 x + 1

. Be careful with your signs!

Check your understanding

3) Factor $2 x^{2} - 3 x - 9$ ‍.

To factor

2 x^{2} - 3 x - 9

, we need to find two integers with a product of

2 \cdot (- 9) = - 18

and a sum of

- 3

To find these numbers, we can create a list of all numbers that multiply to

- 18

. Then, within that list, we look for those that sum to

- 3

Product: $m \cdot n = - 18$ ‍	‍	Sum: $m + n = - 3$ ‍
$(- 1) \cdot 18 = - 18$ ‍		$(- 1) + 18 = 17$ ‍
$1 \cdot (- 18) = - 18$ ‍		$1 + (- 18) = - 17$ ‍
$(- 2) \cdot 9 = - 18$ ‍		$(- 2) + 9 = 7$ ‍
$2 \cdot (- 9) = - 18$ ‍		$2 + (- 9) = - 7$ ‍
$(- 3) \cdot 6 = - 18$ ‍		$(- 3) + 6 = 3$ ‍
$3 \cdot (- 6) = - 18$ ‍		$3 + (- 6) = - 3$ ‍

We can now break the term

- 3 x

into

3 x - 6 x

and use grouping to factor the polynomial:

\begin{aligned} 2 x^{2} - 3 x - 9 \\ = 2 x^{2} + 3 x - 6 x - 9 \\ = (2 x^{2} + 3 x) + (- 6 x - 9) & Group terms \\ = x (2 x + 3) + (- 3) (2 x + 3) & Factor out GCFs \\ = x (2 x + 3) - 3 (2 x + 3) & Simplify \\ = x (2 x + 3) - 3 (2 x + 3) & Common factor! \\ = (2 x + 3) (x - 3) & Factor out 2 x + 3 \end{aligned}

The factored form is

(2 x + 3) (x - 3)

4) Factor $3 x^{2} - 2 x - 5$ ‍.

To factor

3 x^{2} - 2 x - 5

, we need to find two integers with a product of

3 \cdot (- 5) = - 15

and a sum of

- 2

To find these numbers, we can create a list of all numbers that multiply to

- 15

. Then, within that list, we look for those that sum to

- 2

Product: $m \cdot n = - 15$ ‍	‍	Sum: $m + n = - 2$ ‍
$(- 1) \cdot 15 = - 15$ ‍		$(- 1) + 15 = 14$ ‍
$1 \cdot (- 15) = - 15$ ‍		$1 + (- 15) = - 14$ ‍
$(- 3) \cdot 5 = - 15$ ‍		$(- 3) + 5 = 2$ ‍
$3 \cdot (- 5) = - 15$ ‍		$3 + (- 5) = - 2$ ‍

We can now break the term

- 2 x

into

- 5 x + 3 x

and use grouping to factor the polynomial:

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

The factored form is

(3 x - 5) (x + 1)

5) Factor $6 x^{2} - 13 x + 6$ ‍.

To factor

6 x^{2} - 13 x + 6

, we need to find two integers with a product of

6 \cdot 6 = 36

and a sum of

- 13

To find these numbers, we can create a list of all numbers that multiply to

36

. Then, within that list, we look for those that sum to

- 13

Since the numbers sum to a negative, we only need to consider when both factors are negative.

Product: $m \cdot n = 36$ ‍	‍	Sum: $m + n = - 13$ ‍
$(- 1) \cdot (- 36) = 36$ ‍		$(- 1) + (- 36) = - 37$ ‍
$(- 2) \cdot (- 18) = 36$ ‍		$(- 2) + (- 18) = - 20$ ‍
$(- 3) \cdot (- 12) = 36$ ‍		$(- 3) + (- 12) = - 15$ ‍
$(- 4) \cdot (- 9) = 36$ ‍		$(- 4) + (- 9) = - 13$ ‍
$(- 6) \cdot (- 6) = 36$ ‍		$(- 6) + (- 6) = - 12$ ‍

We can now break the term

- 13 x

into

- 4 x - 9 x

and use grouping to factor the polynomial:

\begin{aligned} 6 x^{2} - 13 x + 6 \\ = 6 x^{2} - 4 x - 9 x + 6 \\ = (6 x^{2} - 4 x) + (- 9 x + 6) & Group terms \\ = 2 x (3 x - 2) + (- 3) (3 x - 2) & Factor out GCFs \\ = 2 x (3 x - 2) - 3 (3 x - 2) & Simplify \\ = 2 x (3 x - 2) - 3 (3 x - 2) & Common factor! \\ = (3 x - 2) (2 x - 3) & Factor out 3 x - 2 \end{aligned}

The factored form is

(3 x - 2) (2 x - 3)

When is this method useful?

Well, clearly, the method is useful to factor quadratics of the form

a x^{2} + b x + c

, even when

a \neq 1

However, it's not always possible to factor a quadratic expression of this form using our method.

For example, let's take the expression

2 x^{2} + 2 x + 1

. To factor it, we need to find two integers with a product of

2 \cdot 1 = 2

and a sum of

2

. Try as you might, you will not find two such integers.

Therefore, our method doesn't work for

2 x^{2} + 2 x + 1

, and for a bunch of other quadratic expressions.

It's useful to remember, however, that if this method doesn't work, it means the expression cannot be factored as

(A x + B) (C x + D)

where

A

B

C

, and

D

are integers.

Why is this method working?

Let's take a deep dive into why this method is at all successful. We will have to use a bunch of letters here, but please bear with us!

Suppose the general quadratic expression

a x^{2} + b x + c

can be factored as

(A x + B) (C x + D)

with integers

A

B

C

, and

D

When we expand the parentheses, we obtain the quadratic expression

(A C) x^{2} + (B C + A D) x + B D

Since this expression is equivalent to

a x^{2} + b x + c

, the corresponding coefficients in the two expressions must be equal! This gives us the following relationship between all the unknown letters:

(\underset{a}{\underset{⏟}{A C}}) x^{2} + (\underset{b}{\underset{⏟}{B C + A D}}) x + (\underset{c}{\underset{⏟}{B D}})

Now, let's define

m = B C

and

n = A D

(\underset{a}{\underset{⏟}{A C}}) x^{2} + (\underset{b}{\underset{⏟}{\overset{m}{\overset{⏞}{B C}} + \overset{n}{\overset{⏞}{A D}}}}) x + (\underset{c}{\underset{⏟}{B D}})

According to this definition...

m + n = B C + A D = b

and

m \cdot n = (B C) (A D) = (A C) (B D) = a \cdot c

And so

B C

and

A D

are the two integers we are always looking for when we use this factorization method!

The next step in the method after finding

m

and

n

is to split the

x

-coefficient

(b)

according to

m

and

n

and factor using grouping.

Indeed, if we split the

x

-term

(B C + A D) x

into

(B C) x + (A D) x

, we will be able to use grouping to factor our expression back into

(A x + B) (C x + D)

In conclusion, in this section we...

started with the general expanded expression $a x^{2} + b x + c$ ‍ and its general factorization $(A x + B) (C x + D)$ ‍,
were able to find two numbers, $m$ ‍ and $n$ ‍, such that $m n = a c$ ‍ and $m + n = b$ ‍ $($ ‍we did so by defining $m = B C$ ‍ and $n = A D)$ ‍,
split the $x$ ‍-term $b x$ ‍ into $m x + n x$ ‍, and were able to factor the expanded expression back into $(A x + B) (C x + D)$ ‍.

This process shows why, if an expression can indeed be factored as

(A x + B) (C x + D)

, our method will ensure that we find this factorization.

Thanks for pulling through!

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Sort by:

Alemán
Posted 7 years ago. Direct link to Alemán's post “So, this metod is not app...”
So, this metod is not applicable to this equation: 3x^2+5x+10 for example. What method can I use to factorize this equation? Or there are any quadratics can´t be factorizated?
Button navigates to signup pageComment on Alemán's post “So, this metod is not app...”
(13 votes)
Answer
- David Severin
  Posted 7 years ago. Direct link to David Severin's post “The quadratic formula alw...”
  The quadratic formula always works, so first look at the discriminant b^2 - 4 a c
  For your equation 5*2 - 4(3)(10) or 25 - 120 = - 95
  Since you cannot take SQRT of negative number in real domain
  This cannot be factorable since there is no real solution
  Comment on David Severin's post “The quadratic formula alw...”
  (67 votes)
Vanessa Campos
Posted 6 years ago. Direct link to Vanessa Campos's post “So on number 5, How do yo...”
So on number 5, How do you know to group -4 with 6x^ or -9 with it?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Vicki R
  Posted 2 years ago. Direct link to Vicki R's post “It works either way.”
  It works either way.
  Comment on Vicki R's post “It works either way.”
  (4 votes)
georgefabish
Posted 5 years ago. Direct link to georgefabish's post “On question *4*)factor *3...”
On question 4)factor 3x^2 -2x- 5
In working it out I got the answer (x-1)(3x+5) where as the correct answer was (3x-5)(x+1). The difference appears because there were multiple ways to split the GCF. As far as I can tell they both give the right answer and expand back into 3x^2-2x-5. Is this typical when solving these problems or did I do something wrong or is this just a random special case that happens to work?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “Sorry, but your version c...”
  Sorry, but your version creates the wrong sign on the middle term. You can see this if you multiply your factors. You get: 3x^2+5x-3x-5. Notice: 5x-3x = 2x, not -2x.
  
  Hope this helps.
  Comment on Kim Seidel's post “Sorry, but your version c...”
  (16 votes)
aeppa
Posted 4 years ago. Direct link to aeppa's post “I understand how to facto...”
I understand how to factor the quadratics but I don't really understand the part on how it works.
Button navigates to signup pageComment on aeppa's post “I understand how to facto...”
(11 votes)
Answer
Mansi
Posted 7 years ago. Direct link to Mansi's post “How would I solve a quest...”
How would I solve a question like this: (2(1-x))/3 - 8/1=1/(6x)-(2x-3)/3
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Santiago Fiz
  Posted 3 years ago. Direct link to Santiago Fiz's post “Multiply both sides by 3,...”
  Multiply both sides by 3, which gets -2x-22=(1/2x)-2x-3
  Add 2x, so -22=(1/2x)-3. Then add 3, so -19=(1/2x). Multiply by 2x so -38x=1. Then divide by -38, so
  x = (-1 / 38)
  Button navigates to signup page
  (3 votes)
NIk NIk
Posted 7 years ago. Direct link to NIk NIk's post “Why do we multiple 2 and ...”
Why do we multiple 2 and 3 in the first example?
Button navigates to signup pageComment on NIk NIk's post “Why do we multiple 2 and ...”
(7 votes)
Answer
hari the destroyer
Posted 10 months ago. Direct link to hari the destroyer's post “What if the resulting gro...”
What if the resulting grouping ends up with contradicting signs inside the parenthesis (like 2x(x-4) -5(1+4) )

This is just an example not the actual question I encountered.
Button navigates to signup pageComment on hari the destroyer's post “What if the resulting gro...”
(4 votes)
Answer
- catterpylar
  Posted 9 months ago. Direct link to catterpylar's post “I'm assuming you mean 2x(...”
  I'm assuming you mean 2x(x-4)-5(x+4)
  You just change the sign of the number you are factoring out.
  For example, in the second part, you have -5(x+4). Multiplying it you get -5x-20. To get (x-4) you would factor out 5 instead of -5, to get 5(x-4)
  Now you have 2x(x-4)+5(x-4)
  which then becomes (2x+5)(x-4)
  You can also do this to the first part, 2x(x-4)
  switch 2x to -2x
  then you get -2x(x+4)-5(x+4) then (-2x-5)(x+4).
  Comment on catterpylar's post “I'm assuming you mean 2x(...”
  (3 votes)
Josie Cunningham
Posted 6 years ago. Direct link to Josie Cunningham's post “What is the easiest way h...”
What is the easiest way how to remember how to factor out quadratics?
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(4 votes)
Answer
- 1000050035
  Posted 4 years ago. Direct link to 1000050035's post “There are some songs on y...”
  There are some songs on youtube to help memorize these formulas. I recommend the quadratic formula set to the tune of "Pop Goes the Weasel", because you can use that formula for a lot of different equations, and the song is very memorable. Hope this answers your question :)
  Button navigates to signup page
  (2 votes)
Remedy
Posted a year ago. Direct link to Remedy's post “A question about the firs...”
A question about the first problem. My answer is "(x+2)(3x+4)", but there are no choices in there. Even if the values of a or b are opposite, is the answer the same?
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(1 vote)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “Your answer is correct (y...”
  Your answer is correct (you can always check it by multiplying the 2 binomials). It matches option B in the answer list. Remember, the commutative property of multiplication tells us that 2(3) = 3(2). This property extends to any multiplication problem, including your binomials. (3x+4)(x+2) is the same as (x+2)(3x+4).
  
  Hope this helps.
  Comment on Kim Seidel's post “Your answer is correct (y...”
  (8 votes)
Jay Mack
Posted 3 years ago. Direct link to Jay Mack's post “the letters are really co...”
the letters are really confusing me im getting mixed up on ac and bc and what this and that equals can someone help me pls
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(3 votes)
Answer
- AuroraAlberts
  Posted a year ago. Direct link to AuroraAlberts's post “Breath, focus, and concen...”
  Breath, focus, and concentrate. You can do this!
  Try writing it down and giving the letters different colours.
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Algebra 1

Course: Algebra 1 > Unit 13

What you need to know before taking this lesson

What you will learn in this lesson

Example 1: Factoring 2x2+7x+3‍

Summary

Check your understanding

Example 2: Factoring 6x2−5x−4‍

Check your understanding

When is this method useful?

Why is this method working?

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Example 1: Factoring $2 x^{2} + 7 x + 3$ ‍

Example 2: Factoring $6 x^{2} - 5 x - 4$ ‍