6.1 Greatest Common Factor and Factoring by Grouping
Greatest Common Factor and Factoring by Grouping
Factored form: A number or expression written as a product of factors.
Examples: 2x + 8 = 2(x + 4)
x2 + 5x + 6 = (x + 2)(x + 3)
Greatest common factor (GCF) of a set of terms: A monomial with the greatest coefficient and degree that
evenly divides all of the given terms.
Method
To find the GCF of two or more monomials,
1. Write the prime factorization in exponential form for each monomial.
2. Write the GCF’s factorization by including the prime factors (and variables) common to all the
factorizations, each raised to its smallest exponent in the factorizations.
3. Multiply the factors in the factorization created in step 2.
Note: If there are no common prime factors, then the GCF is 1.
Examples
Find the GCF
[06]
6m4n9,
15mn5
1 1 4 9
2 3 m n , 3151m1n5
GCF = 31m1n5 = 3mn5
[12]
8(m – n), 11(m – n), m – n
23(m – n)1, 111(m – n)1, (m – n)1
GCF = (m – n)1 = m – n
A Method to Find the Prime Factorization of a Constant
Determine how many times two can go into the constant, then the number of threes, fives, sevens, elevens, etc.
Example:
2 ) 420
2 ) 210
3 ) 105
5 ) 35
7)
7
1
There are 2 twos, 1 three, 1 five, 1 seven
420 = 22315171 = 22●3●5●7
Factor a Monomial GCF Out of the Terms of a Polynomial
The method to factor a monomial GCF out of the terms of a polynomial is:
1. Find the GCF of the terms in the polynomial.
2. Rewrite the polynomial as a product of the GCF and the quotient of the polynomial and the GCF.
Polynomial  GCF 
Examples
[20]
Polynomial
GCF
-2x2y2 - 8x3y + 2xy
-21x2y2 - 23x3y1 + 21x1y1
GCF = 21x1y1 = 2xy
2xy(-1xy - 4x2 + 1) or -2xy(xy + 4x2 – 1)
[30]
a( b – 5 ) + 7 ( b – 5 )
a1(b – 5)1 + 71(b – 5)1
GCF = ( b – 5 )1
(b–5)(a–7)
Factor Polynomials by Grouping
Method to factor a four-term polynomial by grouping,
1. Factor out any monomial GCF (other than 1) that is common to all four terms.
2. Group together pairs of terms and factor the GCF out of each pair.
3. If there is a common binomial factor, then factor it out.
4. If there is no common binomial factor, then interchange the middle two terms and repeat the process. If
there is still no common binomial factor, then the polynomial cannot be factored by grouping.
Examples
[50]
15c3 - 5c2d + 6cd2 – 2d3
no common factor in all four terms
[62]
[nib]
5c2 ( 3c – d ) + 2d2 ( 3c – d )
( 3c – d ) ( 5c2 + 2d2 )
12ac + 12cx - 3ac2 - 3c2x
3c ( 4a + 4x - ac - cx )
3c { 4 ( a + x) - c ( a + x ) }
3c ( a + x ) ( 4 – c )
2x3 – y3 – 2x2y + xy2
no common factor in all four terms nor 1st two and last two, rearrange the middle two even then, it won’t always factor
2x3 - 2x2y - y3 + xy2
2x2 ( x – y ) - y2 ( y – x )
note the ( ) terms are conjugates
2x2 ( x – y ) + y2 ( x – y )
( x – y ) ( 2x2 + y2 )