2.2. Advanced Factoring http://www.ck12.org
pair. In this case it is:
( 2 x− 7 )( 3 x+ 4 ) = 6 x^2 + 8 x− 21 x− 28 = 6 x^2 − 13 x− 28
This method can be extremely long and rely heavily on good guessing which is why the algorithm in Example A is
provided and preferable.
An alternative method is using the quadratic formula as a clue even though this is not an equation set equal to zero.
6 x^2 − 13 x− 28
a= 6 ,b=− 13 ,c=− 28
x=−b±
√b (^2) − 4 ac
2 a =^13 ±
√ 169 − 4 · 6 ·− 28
2 · 6 =^1312 ±^29 =^4212 or−^1612 =^72 or−^43
This means that when set equal to zero, this expression is equivalent to
(x− 7
2
)(x+ 4
3)= 0
Multiplying by 2 and multiplying by 3 only changes the left hand side of the equation because the right hand side
will remain 0. This has the effect of shifting the coefficient from the denominator of the fraction to be in front of
thexjust like in Example A.
6 x^2 − 13 x− 28 = ( 2 x− 7 )( 3 x+ 4 )
Example C
Factor the following expression using grouping: 12x^2 + 4 xz+ 3 xy+yz
Solution: Notice that the first two terms are divisible by both 4 andxand the last two terms are divisible byy. First,
factor out these common factors and then notice that there emerges a second layer of common factors. The binomial
( 3 x+z)is now common to both terms and can be factored out just as before.
12 x^2 + 4 xz+ 3 xy+yz= 4 x( 3 x+z)+y( 3 x+z)
= ( 3 x+z)( 4 x+y)Concept Problem Revisited
The sum or difference of terms with matching odd powers can be factored in a precise pattern because when
multiplied out, all intermediate terms cancel each other out.
a^5 +b^5 = (a+b)(a^4 −a^3 b+a^2 b^2 −ab^3 +b^4 )Whenais distributed:a^5 −a^4 b+a^3 b^2 −a^2 b^3 +ab^4
Whenbis distributed:+a^4 b−a^3 b^2 +a^2 b^3 −ab^4 +b^5
Notice all the inside terms cancel:a^5 +b^5
Vocabulary
To factormeans to rewrite a polynomial expression given as a sum of terms into a product of factors.
Guided Practice
- Show howa^3 −b^3 factors by checking the result given in the guidance section.