Factoring by grouping
Just like it says, factoring by grouping means that you will group terms with common factors before factoring.
As you can see, this is done by grouping a pair of terms. Then, factor each pair of two terms. If you did not understand the example above, keep reading as we explain the concept with more examples.
More examples enplaning factoring by grouping
Factor x
^{2} + 5x + 6
The expression x
^{2} + 5x + 6 has three terms right now, so we need to write it with 4 terms
before we can group terms.
5x = 3x + 2x, so x
^{2} + 5x + 6 becomes x
^{2} + 3x + 2x + 6.
Group x
^{2} with 3x and 2x with 6 and then factor each group.
We get
(x
^{2} + 3x) + (2x + 6) = x*(x + 3) + 2*(x + 3) = (x + 3) * (x + 2)
In this example, if you group x
^{2} with 2x and 3x with 6, you will get the same answer. Try doing that.
Notice that there is more than one way we can expand 5x, so different groupings are possible. 5x is also equal to 4x + x, 6x x, 7x2x, 8x3x, and so forth...
However, not all groupings will work!
This brings light to the fact that this way of factoring by grouping can be very tedious sometimes.
Although it is always good to know, it is not always a straightforward
method to factor trinomials.
Example #2:
x
^{2} + 4x + 12
At first, you may be tempted to say that 4x can be equal to: 2x + 2x, or 3x + x, so one of them will work.
Wrong! The right combination is 6x + 2x
So, x
^{2} + 4x + 12 = x
^{2} + 6x + 2x + 12
Group x
^{2} with 6x and 2x with 12
(x
^{2} + 6x) + (2x + 12) = x *(x − 6) + 2 * (x − 6) = (x − 6)*(x + 2)
Example #3:
3y
^{2} + 14y + 8
3y
^{2} + 14y + 8 = 3y
^{2} + 12y + 2y + 8 = (3y
^{2} + 12y) + (2y + 8) = 3y(y + 4) + 2(y + 4)
So, 3y
^{2} + 14y + 8 = (y + 4)(3y + 2)
Example #4:
11x
^{2} + 41x + 12
This problem is very complicated because you have too many choices for things you can add to get 41x.
Some possibilities are:
.....
.....
46x + 5x
45x + 4x
44x + 3x
40x + 1x
39x + 2x
38x + 3x
36x + 4x
.....
.....
It turns out that the right combination is  44x + 3x
There is good news though since there is a technique to use to find the right combination a little bit faster when factoring by grouping.
Do 11 * 12 = 132
Then, find factors of 132 that will add up to 41
The factors are 44 and 3
11x
^{2} + 41x + 12 = 11x
^{2} + 44x + 3x + 12
11x
^{2} + 44x + 3x + 12 = 11x(x − 4) + 3(x − 4) = (x − 4)(11x + 3)
Example #5:
6x
^{2} 26x + 28
6 * 28 = 168
14 + 12 = 26 and 14 * 12 = 168, so the right combination is
6x
^{2} 26x + 28 = 6x
^{2} + 14x + 12x + 28
6x
^{2} + 14x + 12x + 28 = (6x
^{2} + 14x) + (12x + 28)= 2x(3x + 7) + 4(3x + 7)
6x
^{2} 26x + 28 = (3x + 7) * (2x + 4)
Conclusion:
Use factoring by grouping only if you have no other choices!

Oct 20, 21 04:45 AM
Learn how to find the multiplicity of a zero with this easy to follow lesson
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