Factoring by Grouping
Factoring by grouping is a method of factoring that works on four-term polynomials that have a specific pattern to them. The process goes like this:
|Factor:||x3 + 3x2 + 2x + 6|
|1. Rearrange the terms so that the exponents are in decreasing order, if they aren't already.|
|2. Group the first two and the last two terms together.||(x3 + 3x2) + (2x + 6)|
|3. Factor each of the two groups separately. In our example, you can factor an x2 out of the first group and a 2 out of the second.||x2(x + 3) + 2(x + 3)|
|4. Factor the common factor out of the two groups. In our example, both of the groups have an x + 3 in common. That's what we'll factor out.||(x + 3)(x2 + 2)|
|5. Check the two factors to see if they can be factored further. Neither x2 + 2 or x + 3 can be factored further so (x + 3)(x2 + 2) is our final answer.||(x + 3)(x2 + 2)|
Factoring by grouping requires the original polynomial to have a specific pattern that not all four term polynomials will have. If you do the factorization in step three and the two groups don't have a common factor then you need to go back to square one and try a different approach.
Factor x3 + x2 - 3x - 3
The terms are already in descending order so we can start by grouping the first two and the last two terms:
(x3 + x2) + (-3x - 3)
Notice how I handled the minus signs in the second group. It stayed with the 3x and the two groups get added together. Now we can factor each of the two groups to get
x2(x + 1) - 3(x + 1)
Watch those negative signs again. I factored a -3 out of the second group so both of the terms inside it became positive. I could've written the second part of the expression as + (-3)(x + 1) but it's a little simpler to write the "plus negative three" as subtraction.
Now I can factor the x + 1 out of both parts and get the final factored version:
(x + 1)(x2 - 3)
Factor 2x5 - x4 + 2x2 - x
The terms are already in descending order so we'll start by grouping them
(2x5 - x4) + (2x2 - x)
and then factor each group.
x4(2x - 1) + x(2x - 1)
Now we can factor out the 2x - 1 that both groups have in common go get
(2x - 1)(x4 + x)
At this point, you might be tempted to stop but remember that there's one more step on our procedure list. In this case, the second factor, x4 + x, can be factored further by going back to our discussion about factoring out the greatest common factor. In this case, the greatest common factor of x4 and x is x so we can factor that out to get the final factorization.
(2x - 1)(x · x3 + x · 1)
x(2x - 1)(x3 + 1)
Factor -12x2 + x + 3x3 - 4
In this example, the terms are otu of order so we need to start by rearranging them.
3x3 - 12x2 + x - 4
Notice how the negative sign stayed with then -12 but got written as subtraction. Now we can group the first and laste apri of terms.
(3x3 - 12x2) + (x - 4)
Now we can factor the first and last groups.
3x2(x - 4) + (x - 4)
At this point, the second group might be a little confusing. When we factor out the x - 4, what's going to be left? Keep in mind that factoring never causes a term to "go away" so something has to be left in that spot. In this case, it's going to be a 1. To make that clear, we can explicitly write the 1 in front of that term like this:
3x2(x - 4) + 1 (x - 4)
Now we can factor out the x - 4 to get our final factored version.
(x - 4)(3x2 + 1)
Factor x3 + 2x2 - 4x - 8
The factrs in the polynomial are already in order so we can move right to the grouping step.
(x3 + 2x2) + (-4x - 8)
Notice how I kept the - sign with the four and added the two groups together. Now, since both of the terms in the second group are negative, I'll factor a -4 out of it.
x2(x + 2) - 4(x + 2)
Now I can factor out the x + 2 out to get
(x2 - 4)(x + 2)
Now, if you look closely at x2 - 4, you'll see that it can be factored further using the techniques from the "Factoring Trinomials 1" lesson.
(x - 2)(x + 2)(x + 2)
If you really wanted to streamline the answer, the last two factors could be combined together to give us
(x - 2)(x + 2)2
Directions: This solution has 4 steps. To see a description of each step click on the boxes on the left side below. To see the calculations, click on the corresponding box on the right side. Try working out the solution yourself and use the descriptions if you need a hint and the calculations to check your solution.