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Factoring by Grouping

Use associative and commutative properties with factoring

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Factoring by Grouping

What if you had a polynomial expression like \begin{align*}3x^2 - 6x + 2x - 4\end{align*}3x26x+2x4 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 \begin{align*}2x+2y+ax+ay\end{align*}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 \begin{align*}a\end{align*}a. Factor 2 from the first two terms and factor \begin{align*}a\end{align*}a from the last two terms:

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

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

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

Example B

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

Solution

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

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

Now factor \begin{align*}(x+2)\end{align*}(x+2) from both terms: \begin{align*}(x+2)(3x+4)\end{align*}(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 \begin{align*}ax^2+bx+c\end{align*}ax2+bx+c, where \begin{align*}a \neq 1\end{align*}a1.

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

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

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

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

Solution:

Let’s follow the steps outlined above:

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

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

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

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

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

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

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

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

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

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

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

Step 1: \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:

\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: \begin{align*}-11x = -3x - 8x\end{align*}, so the problem becomes:

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

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

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

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

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

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

Solution:

Let’s follow the steps outlined above:

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

Step 1: \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:

\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: \begin{align*}-6x = -x - 5x\end{align*}, so the problem becomes:

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

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

\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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