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The following polynomial has four terms:
Notice that there is no common factor among the four terms (no GCF).
However the first two terms do have a common factor of ๐ฆ and the last
two terms have a common factor of 3. So while we canโt factor the
polynomial by taking out a GCF, we can factor by grouping. This means
grouping the first two terms and factoring out a GCF, then grouping the
last two terms and factoring out a GCF.
We now have a sum of two terms, and both terms have a common factor
of (๐ฅ + 2 ). If we take out the GCF of (๐ฅ + 2 ) we are left with the
following:
This is an example of factoring a polynomial by grouping the terms.
Factor by grouping:
- grouping the terms of a polynomial and factoring out a GCF from
each group
- you can group any terms that have a common factor
o this means the order of the two middle terms can be reversed
and the final factored answer will remain the same; this will be
important on the next page when we use the ๐๐-method to factor
3
2
3
2
3
2
3
2
2
2
2
๐
๐
Example 1: Factor the following polynomials by grouping.
a. ๐ฅ
3
2
3
2
Always check to see if
the terms in the polynomial
have a GCF. In this problem,
the four terms do not have a
GCF, so I will simply factor
by grouping.
3
2
2
๐
b. 6 ๐ฅ
4
3
2
c. 16 ๐ฅ
2
To factor this trinomial using the ๐๐-method, I would start by trying to
find two numbers with a of product ๐๐
and a sum of ๐
( 24 ). In this case, those two numbers are 12 and 12 , so I will replace
24 ๐ฅ with 12 ๐ฅ + 12 ๐ฅ in order to then factor by grouping.
2
Since I end up with the same binomial twice, I can express it as a perfect
square.
๐
Think
about
the
signs of
the
product
and the
sum.
1 , 6
โ 1 , โ 6
2 , 3
โ 2 , โ 3
โ 5 , 12
โ 6 , 10
When to factor out a negative factor rather than a
positive factor:
There are two scenarios in which it is beneficial to factor out a negative
factor rather than a positive factor:
- When the leading coefficient is negative
2
2
2
The trinomial ๐ฅ
2
โ 5 ๐ฅ โ 6 should be easier to factor than โ๐ฅ
2
which we would have had if weโd factored out 3 instead of โ 3.
- To make the binomials match when factoring by grouping
2
2
Had I factored out 6 from โ 6 ๐ฅ โ 6 , I would have been left with the
binomial (โ๐ฅ โ 1 ) which would not have matched (๐ฅ + 1 ). By factoring
out a โ 6 instead, I had a common factor of
, which I was then able
to factor out.
Answers to Examples:
1 a.
2
; 1 b. 2
2
2 a.
; 2 b. ๐ฅ
2
; 2 c.
2
3 a. โ 1 (๐ฅ โ 5 )(๐ฅ + 30 ) ; 3b. โ๐ฅ
