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# 9.13: Factoring by Grouping

Difficulty Level: At Grade Created by: CK-12
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What if you had a polynomial expression like 3x26x+2x4\begin{align*}3x^2 - 6x + 2x - 4\end{align*} 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\begin{align*}2x+2y+ax+ay\end{align*}.

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\begin{align*}a\end{align*}. Factor 2 from the first two terms and factor a\begin{align*}a\end{align*} from the last two terms:

2x+2y+ax+ay=2(x+y)+a(x+y)\begin{align*}2x + 2y + ax + ay = 2(x + y) + a(x + y)\end{align*}

Now we notice that the binomial (x+y)\begin{align*}(x + y)\end{align*} is common to both terms. We factor the common binomial and get:

(x+y)(2+a)\begin{align*}(x + y)(2 + a)\end{align*}

#### Example B

Factor 3x2+6x+4x+8\begin{align*}3x^2+6x+4x+8\end{align*}.

Solution

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

3x(x+2)+4(x+2)\begin{align*}3x(x+2)+4(x+2)\end{align*}

Now factor (x+2)\begin{align*}(x+2)\end{align*} from both terms: (x+2)(3x+4)\begin{align*}(x+2)(3x+4)\end{align*}.

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 ax2+bx+c\begin{align*}ax^2+bx+c\end{align*}, where a1\begin{align*}a \neq 1\end{align*}.

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

To factor a quadratic polynomial where a1\begin{align*}a \neq 1\end{align*}, we follow these steps:

1. We find the product ac\begin{align*}ac\end{align*}.
2. We look for two numbers that multiply to ac\begin{align*}ac\end{align*} and add up to b\begin{align*}b\end{align*}.
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) 3x2+8x+4\begin{align*}3x^2+8x+4\end{align*}

b) 6x211x+4\begin{align*}6x^2-11x+4\end{align*}

Solution:

Let’s follow the steps outlined above:

a) 3x2+8x+4\begin{align*}3x^2+8x+4\end{align*}

Step 1: ac=34=12\begin{align*}ac = 3 \cdot 4 = 12\end{align*}

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

121212=112=26=34andandand1+12=132+6=8This is the correct choice.3+4=7\begin{align*}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\end{align*}

Step 3: Re-write the middle term: 8x=2x+6x\begin{align*}8x = 2x + 6x\end{align*}, so the problem becomes:

3x2+8x+4=3x2+2x+6x+4\begin{align*}3x^2+8x+4=3x^2+2x+6x+4\end{align*}

Step 4: Factor an x\begin{align*}x\end{align*} from the first two terms and a 2 from the last two terms:

x(3x+2)+2(3x+2)\begin{align*}x(3x+2)+2(3x+2)\end{align*} Now factor the common binomial (3x+2)\begin{align*}(3x + 2)\end{align*}:

(3x+2)(x+2)This is the answer.\begin{align*}(3x+2)(x+2) \qquad This \ is \ the \ answer.\end{align*}

To check if this is correct we multiply (3x+2)(x+2)\begin{align*}(3x+2)(x+2)\end{align*}:

3x+2x+2  6x+43x2+2x3x2+8x+4\begin{align*}& \qquad \ \ 3x+2\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;x+2\;}\\ & \quad \quad \ \ 6x+4\\ & \underline{3x^2+2x \;\;\;\;\;}\\ & 3x^2+8x+4\end{align*}

The solution checks out.

b) 6x211x+4\begin{align*}6x^2-11x+4\end{align*}

Step 1: ac=64=24\begin{align*}ac = 6 \cdot 4 = 24\end{align*}

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

2424242424242424=124=1(24)=212=2(12)=38=3(8)=46=4(6)andandandandandandandand1+24=251+(24)=252+12=142+(12)=143+8=113+(8)=11(Correct choice)4+6=104+(6)=10\begin{align*}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\end{align*}

Step 3: Re-write the middle term: 11x=3x8x\begin{align*}-11x = -3x - 8x\end{align*}, so the problem becomes:

6x211x+4=6x23x8x+4\begin{align*}6x^2-11x+4=6x^2-3x-8x+4\end{align*}

Step 4: Factor by grouping: factor a 3x\begin{align*}3x\end{align*} from the first two terms and a -4 from the last two terms:

3x(2x1)4(2x1)\begin{align*}3x(2x-1)-4(2x-1)\end{align*}

Now factor the common binomial (2x1)\begin{align*}(2x - 1)\end{align*}:

(2x1)(3x4)This is the answer.\begin{align*}(2x-1)(3x-4) \qquad This \ is \ the \ answer.\end{align*}

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 5x26x+1\begin{align*}5x^2-6x+1\end{align*} by grouping.

Solution:

Let’s follow the steps outlined above:

5x26x+1\begin{align*}5x^2-6x+1\end{align*}

Step 1: ac=51=5\begin{align*}ac = 5 \cdot 1 = 5\end{align*}

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

55=15=1(5)andand1+5=61+(5)=6(Correct choice)\begin{align*}5 &= 1 \cdot 5 && \text{and} && 1 + 5 = 6\\ 5 &= -1 \cdot (-5) && \text{and} && -1 + (-5) = -6 \qquad (Correct \ choice)\end{align*}

Step 3: Re-write the middle term: 6x=x5x\begin{align*}-6x = -x - 5x\end{align*}, so the problem becomes:

5x26x+1=5x2x5x+1\begin{align*}5x^2-6x+1=5x^2-x-5x+1\end{align*}

Step 4: Factor by grouping: factor an x\begin{align*}x\end{align*} from the first two terms and a1\begin{align*}a - 1\end{align*} from the last two terms:

x(5x1)1(5x1)\begin{align*}x(5x-1)-1(5x-1)\end{align*}

Now factor the common binomial \begin{align*}(5x - 1)\end{align*}:

\begin{align*}(5x-1)(x-1) \qquad This \ is \ the \ answer.\end{align*}

### Practice

Factor by grouping.

1. \begin{align*}6x^2-9x+10x-15\end{align*}
2. \begin{align*}5x^2-35x+x-7\end{align*}
3. \begin{align*}9x^2-9x-x+1\end{align*}
4. \begin{align*}4x^2+32x-5x-40\end{align*}
5. \begin{align*}2a^2-6ab+3ab-9b^2\end{align*}
6. \begin{align*}5x^2+15x-2xy-6y\end{align*}

Factor the following quadratic trinomials by grouping.

1. \begin{align*}4x^2+25x-21\end{align*}
2. \begin{align*}6x^2+7x+1\end{align*}
3. \begin{align*}4x^2+8x-5\end{align*}
4. \begin{align*}3x^2+16x+21\end{align*}
5. \begin{align*}6x^2-2x-4\end{align*}
6. \begin{align*}8x^2-14x-15\end{align*}

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Color Highlighted Text Notes

### Vocabulary Language: English

Factor by Grouping

Factoring by grouping is a method of factoring a polynomial by factoring common monomials from groups of terms.

Grouping Symbols

Grouping symbols are parentheses or brackets used to group numbers and operations.

Volume

Volume is the amount of space inside the bounds of a three-dimensional object.

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