I feel incredibly stupid, but due to absence I missed my teacher's explanation and do not understand. The objective of my assignment is to factor a polynomial by grouping terms. For example, a(a-b)+4b(a-b)-a(a-b).. I would really appreciate any instruction anyone can offer!
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Factoring by Grouping
- Thread starter Lacus
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gcfan4life1023
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- Oct 11, 2005
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i can help you out if you give me an example of a problem that you need to solve
Lacus said:I feel incredibly stupid, but due to absence I missed my teacher's explanation and do not understand. The objective of my assignment is to factor a polynomial by grouping terms. For example, a(a-b)+4b(a-b)-a(a-b).. I would really appreciate any instruction anyone can offer!
"Factoring by grouping" is really an application of using the Distributive Property to remove a common factor.
Suppose you need to factor this expression:
4ab + 3b
You'd look for a common factor, and notice that each term has a factor of b. Put the "b" outside a set of parentheses and what is left from each term inside the parentheses:
b(4a + 3)
Now, suppose your expression had been this, instead:
4a(x + y) + 3(x + y)
We still have a common factor, but this time, the common factor is (x + y). We'll do the same thing: we will put the common factor outside a set of parentheses, and what is left from each term inside the parentheses. But, because the common factor is itself a binomial, we will need to put it inside its own set of parentheses:
(x + y)(4a + 3)
How about if we had started with this?
4ax + 4ay + 3x + 3y
Think of the first two terms as a "group". Do the terms in this group have a common factor? Yes, they do....4a is common to both of these terms, so we'll remove it:
4a(x + y) + 3x + 3y
Next, look at the last two terms to see if they have a common factor, and they do....each term has a factor of 3. Remove the common factor from these terms:
4a(x + y) + 3(x + y)
And we have arrived at the "starting point" of the previous example. These terms have a common factor of (x + y), which we will remove to get
(x + y)(4a + 3)
And the expression has been factored.
Here's one more example. Factor
2x<SUP>3</SUP> - 6x<SUP>2</SUP> + 5x - 15
Think of the first two terms as a group. Do they have a common factor? Yes, each term has a factor of 2x<SUP>2</SUP>. Remove this common factor from the first two terms:
2x<SUP>2</SUP>(x - 3) + 5x - 15
Next, look at the last two terms. They each have a factor of 5, so we'll remove a 5 from this group:
2x<SUP>2</SUP>(x - 3) + 5(x - 3)
Do you see that we now have two terms, and that (x - 3) is a common factor?
2x<SUP>2</SUP>(x - 3) + 5(x - 3)
So our last step is to remove the common factor of (x - 3) and put what is left from each term inside a separate set of parentheses:
(x - 3)(2x<SUP>2</SUP> + 5)
We need to look at (2x<SUP>2</SUP> + 5) to see if it can be factored any further (sometimes, you'll get an expression inside one set of parentheses which is a difference of two squares, or something else which can be factored). In this case, we've got something that cannot be broken down into simpler factors, so we are done. The complete factorization is
(x - 3)(2x<SUP>2</SUP> + 5)
Remember that you can check a factorization by multiplying the factors together; you should get the expression you started with.
If you would like further explanation of this method, you might want to look at some of the modules at http://www.purplemath.com.