Factoring by Grouping

It may be 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. The following example illustrates how this process works.

Example 3

Factor \(\begin{align*}2x+2y+ax+ay\end{align*}\).

Solution

There isn't a common factor for all four terms in this example. However, there is a factor of 2 that is common to the first two terms and there is a factor of \(\begin{align*}a\end{align*}\) that is common to the last two terms. Factor 2 from the first two terms and factor  \(\begin{align*}a\end{align*}\) from the last two terms.

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

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

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

Our polynomial is now factored completely.

We know how to factor Quadratic Trinomials \(\begin{align*}(ax^2+bx+c)\end{align*}\) where  \(\begin{align*}a \neq 1\end{align*}\) using methods we have previously learned. To factor a quadratic polynomial where \(\begin{align*}a \neq 1\end{align*}\), we follow the following steps.

  1. We find the product \(\begin{align*}ac\end{align*}\).
  2. We look for two numbers that multiply to give \(\begin{align*}ac\end{align*}\) and add to give \(\begin{align*}b\end{align*}\).
  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 4

Factor \(\begin{align*}3x^2+8x+4\end{align*}\) by grouping.

Solution:

Follow the steps outlined above.

\(\begin{align*}ac=3 \cdot 4=12\end{align*}\)

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

\(\begin{align*}12&=1 \times 12 && and && 1+12=13\\ 12 & =2 \times 6 && and && 2+6=8 \qquad \text{This is the correct choice}.\end{align*}\)

Rewrite the middle term as: \(\begin{align*}8x=2x+6x\end{align*}\), so the problem becomes the following.

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

Factor an \(\begin{align*}x\end{align*}\) from the first two terms and 2 from the last two terms.

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

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

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

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

In this example, all the coefficients are positive. What happens if the \(\begin{align*}b\end{align*}\) is negative?

Example 5:

Factor \(\begin{align*}6x^2-11x+4\end{align*}\) by grouping.

Solution:

\(\begin{align*}ac=6 \cdot 4 = 24\end{align*}\)

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

\(\begin{align*}24&=1\times 24 && and && 1+24=25\\ 24&=(-1) \times (-24) && and && (-1)+(-24)=-25\\ 24&=2 \times 12 && and && 2+12=14\\ 24&=(-2) \times (-12) && and && (-2)+(-12)=-14\\ 24&=3 \times 8 && and && 3+8=11\\ 24&=(-3) \times (-8) && and && (-3)+(-8)=-11 \quad \text{This is the correct choice}.\end{align*}\)

Rewrite the middle term as \(\begin{align*}-11x=-3x-8x\end{align*}\), so the problem becomes:

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

Factor by grouping. Factor a \(\begin{align*}3x\end{align*}\) from the first two terms and factor –4 from the last two terms.

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

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

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

Back to '9.2: Factoring Polynomials by Grouping'