Factoring is the process of undoing multiplication of two binomials. The procedure is the same regardless of if \(a=1\) or \(a\ne1\).
There are two parts to the GCF: #1 the largest factor of all the coefficients or each term; #2 the largest number of variables in each term. To be efficient, the need for the greatest common factor is critical. Many times there is an easy common factor, but the GCF requires the greatest one. A quick example with just two terms will show how this works.
Find the GCF of \(24x^6-56x^3+16x^2\).
| First, handle the coefficients of \(24,\;-56\) and \(16\) by finding the largest factor common to all: | \(\begin{array}{l} 24=\color{blue}{8}\cdot3 \\ -56=-1\cdot\color{blue}{8}\cdot7 \\ 16=\color{blue}{8}\cdot2 \end{array}\) |
| The GCF of the coefficents is \(8\), since the left over factors of \(3,\;7\) and \(2\) only have a common factor of \(1\). | |
| ALTERNATE APPROACH: A common factor is \(2\). This gives: |
\(\begin{array}{l} 24=2\cdot12 \\ -56=-1\cdot2\cdot28 \\ 16=2\cdot8 \end{array}\) |
| However, the left over multipliers of \(12,\;28\) and \(8\) all have another GCF of \(2\). This gives: | \(\begin{array}{l} 24=2\cdot2\cdot6 \\ -56=-1\cdot2\cdot2\cdot14 \\ 16=2\cdot2\cdot4 \end{array}\) |
| Again, the left over multipliers of \(6,\;14\) and \(4\) have a GCF of \(2\). Repeating the steps: | \(\begin{array}{l} 24=2\cdot2\cdot2\cdot3\implies\color{blue}{8}\cdot3 \\ -56=-1\cdot2\cdot2\cdot2\cdot7\implies\color{blue}{8}\cdot7 \\ 16=2\cdot2\cdot2\cdot2\implies\color{blue}{8}\cdot2 \end{array}\) |
| Now the left over multipliers have a GCF of \(1\) and a common multiplier of \(2\cdot2\cdot2=8\), making the GCF of the coefficients \(8\). This technique shows how you can find a GCF by pulling off one common factor at a time. | |
| For the variable part, the exponent of \(6\) means there are \(6\;x\)-variables in term one, \(3\;x\)-variables in term two and only \(2\;x\)-variables in the last term. The largest number of \(x\)-variables in each group is two, or \(x^2\). This is the last portion to make up the GCF, hence the overall GCF of the three terms is \(8x^2\). | |
The purpose here is to find the two binomials that multiply together and make the trinomial: \(ax^2+bx+c=(a_1x+d_1)(a_2x+d_2)\). The following example will go through the procedure.
Factor \(3x^2-16x-12\) into the two binomial components.
| Step | Description | Example |
|---|---|---|
| #1: | Multiply first and last coefficients \(a\cdot c\): | \(a=3,\;c=-12\implies a\cdot c=-36\) |
| #2: | Find factors of \(a\cdot c\) that ADD up to \(b\): | Factors of \(-36\) that ADD up to \(-16\): \( \begin{array}{c} 1 & \color{blue}{2} & 3 & 4 & 6 \\ -36 & \color{blue}{-18} & -12 & -9 & -6 \end{array}\) |
| Note: Pay close attention to positives and negatives here. When \(ac\) is negative, one factor must be positive and the other negative. The largest factor will have the same sign as \(b\). When \(ac\) is positive, then both factors are positive or both factors are negative, same sign as \(b\). | ||
| #3: | Rewrite the trinomial into four terms using the factors found: | \(3x^2-18x+2x-12\) |
| #4: | Separate the first two terms from the last two terms into groups. Then take out the GCF of each group. This is called Factor by Grouping. | \(\color{blue}{\left(3x^2-18x\right)}+\color{blue}{\left(2x-12\right)}\) \(\implies 3x(x-6)+2(x-6)\) |
| #5: | If Step#4 is done correctly, a common binomial exists, then the GCFs form the other binomial: | \((x-6)(3x+2)\) or \((3x+2)(x-6)\color{green}{\checkmark}\) |
The video below shows the same example worked above, but using the Box, or Area, Model. It is just another way of organizing all the GCF work.
Now that you have seen both methods of factoring, it is time to practice this skill. Below are 4 problems, all factoring. I encourage you to attempt first, then check your solutions with mine.
Consider the following: \(16x^2-9\). At first, this problem looks completely different. It does not follow the trinomials form above, the middle term \(bx\) is missing. However, instead of missing, consider the coefficient of a \(0\): \(16x^2-9\rightarrow16x^2+0x-9\). Now it is a trinomial and normal factoring methods can be applied: \(16\cdot-9=-144\), now find factors that ADD up to \(0\). The factors here that work are \(-12\) and \(12\), using Factoring by Grouping: $$(16x^2-12x)+(12x-9)\implies4x(4x-3)+3(4x-3)\implies(4x-3)(4x+3)\color{green}{\checkmark}$$
This sort of problem has a shortcut by recognizing there are perfect square numbers involved. Also, the problem is subtraction, which is why it is called a Difference of Squares. Note that, $$16x^2-9\rightarrow\left(4x\right)^2-(3)^2$$ and the perfect square factors of \(4x\) and \(3\) are exactly the factors found in the grouping method. Once these factors are found, one binomial is the sum of the factors and the other binomial is the difference: $$(4x+3)(4x-3)$$
I showed the shortcut after the longer grouping method because I want to impress that the factoring procedure will ALWAYS factor these trinomials. Shortucts sometimes lose their luster if we don't understand the longer method first.