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Factoring by Grouping

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Factoring by Grouping

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.

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CK-12 Foundation: 0913S Factoring 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.

CK-12 Foundation: Factoring By Grouping

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

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