<meta http-equiv="refresh" content="1; url=/nojavascript/"> Factoring by Grouping ( Read ) | Algebra | CK-12 Foundation

# Factoring by Grouping

%
Best Score
Practice Factoring by Grouping
Best Score
%
Factoring by Grouping
0  0  0

What if you had a polynomial expression like $3x^2 - 6x + 2x - 4$ in which some of the terms shared a common factor but not all of them? How could you factor this expression? After completing this Concept, you'll be able to factor polynomials like this one by grouping.

### Guidance

Sometimes, we can factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factor by grouping.

The next example illustrates how this process works.

#### Example A

Factor $2x+2y+ax+ay$ .

Solution

There is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of $a$ . Factor 2 from the first two terms and factor $a$ from the last two terms:

$2x + 2y + ax + ay = 2(x + y) + a(x + y)$

Now we notice that the binomial $(x + y)$ is common to both terms. We factor the common binomial and get:

$(x + y)(2 + a)$

#### Example B

Factor $3x^2+6x+4x+8$ .

Solution

We factor 3x from the first two terms and factor 4 from the last two terms:

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

Now factor $(x+2)$ from both terms: $(x+2)(3x+4)$ .

Now the polynomial is factored completely.

Factor Quadratic Trinomials Where a ≠ 1

Factoring by grouping is a very useful method for factoring quadratic trinomials of the form $ax^2+bx+c$ , where $a \neq 1$ .

A quadratic like this doesn’t factor as $(x \pm m)(x \pm n)$ , so it’s not as simple as looking for two numbers that multiply to $c$ and add up to $b$ . Instead, we also have to take into account the coefficient in the first term.

To factor a quadratic polynomial where $a \neq 1$ , we follow these steps:

1. We find the product $ac$ .
2. We look for two numbers that multiply to $ac$ and add up to $b$ .
3. We rewrite the middle term using the two numbers we just found.
4. We factor the expression by grouping.

Let’s apply this method to the following examples.

#### Example C

Factor the following quadratic trinomials by grouping.

a) $3x^2+8x+4$

b) $6x^2-11x+4$

Solution:

Let’s follow the steps outlined above:

a) $3x^2+8x+4$

Step 1: $ac = 3 \cdot 4 = 12$

Step 2: The number 12 can be written as a product of two numbers in any of these ways:

$12 &= 1 \cdot 12 && \text{and} && 1 + 12 = 13\\12 &= 2 \cdot 6 && \text{and} && 2 + 6 = 8 \qquad This \ is \ the \ correct \ choice.\\12 &= 3 \cdot 4 && \text{and} && 3 + 4 = 7$

Step 3: Re-write the middle term: $8x = 2x + 6x$ , so the problem becomes:

$3x^2+8x+4=3x^2+2x+6x+4$

Step 4: Factor an $x$ from the first two terms and a 2 from the last two terms:

$x(3x+2)+2(3x+2)$ Now factor the common binomial $(3x + 2)$ :

$(3x+2)(x+2) \qquad This \ is \ the \ answer.$

To check if this is correct we multiply $(3x+2)(x+2)$ :

$& \qquad \ \ 3x+2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x+2\;}\\& \quad \quad \ \ 6x+4\\& \underline{3x^2+2x \;\;\;\;\;}\\& 3x^2+8x+4$

The solution checks out.

b) $6x^2-11x+4$

Step 1: $ac = 6 \cdot 4 = 24$

Step 2: The number 24 can be written as a product of two numbers in any of these ways:

$24 &= 1 \cdot 24 && \text{and} && 1 + 24 = 25\\24 &= -1 \cdot (-24) && \text{and} && -1 + (-24) = -25\\24 &= 2 \cdot 12 && \text{and} && 2 + 12 = 14\\24 &= -2 \cdot (-12) && \text{and} && -2 + (-12) = -14\\24 &= 3 \cdot 8 && \text{and} && 3 + 8 = 11\\24 &= -3 \cdot (-8) && \text{and} && -3 + (-8) = -11 \qquad (Correct \ choice) \\24 &= 4 \cdot 6 && \text{and} && 4 + 6 = 10\\24 &= -4 \cdot (-6) && \text{and} && -4 + (-6) = -10$

Step 3: Re-write the middle term: $-11x = -3x - 8x$ , so the problem becomes:

$6x^2-11x+4=6x^2-3x-8x+4$

Step 4: Factor by grouping: factor a $3x$ from the first two terms and a -4 from the last two terms:

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

Now factor the common binomial $(2x - 1)$ :

$(2x-1)(3x-4) \qquad This \ is \ the \ answer.$

Watch this video for help with the Examples above.

### Vocabulary

• It is possible to factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called factoring by grouping .

### Guided Practice

Factor $5x^2-6x+1$ by grouping.

Solution:

Let’s follow the steps outlined above:

$5x^2-6x+1$

Step 1: $ac = 5 \cdot 1 = 5$

Step 2: The number 5 can be written as a product of two numbers in any of these ways:

$5 &= 1 \cdot 5 && \text{and} && 1 + 5 = 6\\5 &= -1 \cdot (-5) && \text{and} && -1 + (-5) = -6 \qquad (Correct \ choice)$

Step 3: Re-write the middle term: $-6x = -x - 5x$ , so the problem becomes:

$5x^2-6x+1=5x^2-x-5x+1$

Step 4: Factor by grouping: factor an $x$ from the first two terms and $a - 1$ from the last two terms:

$x(5x-1)-1(5x-1)$

Now factor the common binomial $(5x - 1)$ :

$(5x-1)(x-1) \qquad This \ is \ the \ answer.$

### Practice

Factor by grouping.

1. $6x^2-9x+10x-15$
2. $5x^2-35x+x-7$
3. $9x^2-9x-x+1$
4. $4x^2+32x-5x-40$
5. $2a^2-6ab+3ab-9b^2$
6. $5x^2+15x-2xy-6y$

Factor the following quadratic trinomials by grouping.

1. $4x^2+25x-21$
2. $6x^2+7x+1$
3. $4x^2+8x-5$
4. $3x^2+16x+21$
5. $6x^2-2x-4$
6. $8x^2-14x-15$