Factoring by Grouping
April 19, 2025
In our previous lesson, we explored some common techniques in factoring. Now, we’re going to learn a new and interesting method, which is factoring by grouping. Let’s dive in and see how this unique technique can make factoring easier and more manageable!
A quadratic polynomial expression expressed in general form is the equation 𝑎2 + bx + c where, 𝑎, b, and c are real numbers with 𝑎 ≠ 0.
We will be focusing on factoring such expressions with 𝑎 ≠ 1. Here are the steps:
- Find the product 𝑎c.
- List all the possible factors of ac, and look for the specific pair of numbers that gives a sum of b.
- Afterwards, we will split b as the sum of those found numbers in the second step.
- Lastly, we factor out the expression by the common binomial.
Factor 6x2 + 7x + 1
Note that in the expression 6x^2 + 7x + 1, we obtain a = 6 \space \text {and} \space c = 1. \space \text {Thus,} \space a = 6 \space \text {and} \space c = 1. \space \text {Thus,} \space ac = (6)(1) = 6. Now that we identified 𝑎c = 6, we then list all of its factors:
By inspection, we can see that the sum of 1 and 6 is 7, which is equal to b. Hence, we split 7x as
With this, we can rewrite 6x2 + 7x+1 as
We can now perform factoring by grouping:
Therefore,
Factor 3n2 - 8n + 4
Note that in the expression 3n2 − 8n + 4, we obtain 𝑎 = 3 and b = 4. Thus, 𝑎c = (3)(4) = 12. Now that we identified 𝑎c = 12, we then list all of its factors:
By inspection, we can see that the sum of −2 and −6 is −8, which is equal to b. Hence, we split −8n as
With this, we can rewrite 3n2− 8n + 4 as 3n2− 2n − 6n + 4. We can now perform factoring by grouping:
Therefore,
Factor 2x2 + 9x - 5
Note that in the expression 2x^2 + 9x - 5, we obtain a = 2 \space \text{and} \space c = -5. Thus, ac = (2)(-5) = -10.
Now that we identified ac = -10, we then list all of its factor pairs:
By inspection, we can see that the sum of -1 \text{ and } 10 \text{ is } 9, which is equal to b. Hence, we split 9x \text{ as }
While factoring by grouping works effectively on general quadratic trinomials, it is important to note that if there are no factor pair matches b, then the said expression is not factorable.
This simple steps will make sure that the factored form such expressions is obtained:
- Group the terms into two pairs. (Usually, you can group the first two terms together and the last two terms together.)
- Factor out the greatest common factor (GCF) from each group.
- If both groups now have the same binomial factor, write the expression as a product of two factors.
Factor 40xy + 30x - 100y - 75
We start by grouping the terms in pairs. Any pair will work, but for the sake of simplicity, we will group the first two terms together and the last two terms together:
Now that the terms are paired, we can now perform factoring by grouping:
\text{Therefore, } 40xy + 30x - 100y - 75 = \boxed{5(2x - 5)(4y + 1)}.
Factoring 140ab - 60a2 + 168b - 72a
We start by grouping the terms in pairs. For simplicity, we group the first two terms together and the last two terms together:
Now that the terms are paired, we can now perform factoring by grouping:
\text{Therefore,} \space 140ab - 60a^2 + 168b - 72a = \boxed{4(5a + 6) (7b -3a)}