Download M098 Carson Elementary and Intermediate Algebra 3e Section 6.1 Objectives

M098
Carson Elementary and Intermediate Algebra 3e
Section 6.1
Objectives
1.
2.
3.
4.
List all possible factors for a given number.
Find the greatest common factor of a set of number or monomials.
Write a polynomial as a product of a monomial GCF and a polynomial.
Factor by grouping.
Vocabulary
Factored Form
Product
Natural Number
Greatest Common
Factor
Prime factorization
Prime number
A number or expression written as a product of factors.
Answer to a multiplication problem
Counting numbers {1, 2, 3, …}
The largest natural number that divides all given numbers with no remainder.
A factorization that contains only prime numbers.
A number that is only divisible by 1 and itself.
Prior Knowledge:
1.
Factors are numbers that are being multiplied and terms are numbers that are being added or
subtracted.
2(3)
2x
5xy
x+y
3(x + y)
(x + y)(3x + 4)(2x + 1)
2 factors
2 factors
3 factors
0 factors, 2 terms
2 factors, the second factor has 2 terms
3 factors each with 2 terms
New Concepts
1. Factored form means the quantities are multiplied together.
Example 1:
3(x + 5)
3x + 15
(3x + 2)(x + 5)
3x(x + 5) + 2(x + 5)
Factored form
Not factored form
Factored form
Not factored form
2. Find the greatest common factor.
Finding the GCF of the coefficients is the same process that is used to reduce fractions. Look for the
largest number that divides into each coefficient.
When the numbers are small, the GCF can usually be found mentally, but when the numbers are larger,
use either the listing method or the prime factorization method.

Listing Method: List all of the possible factors for each given number and then find the largest
factor that is common to all lists.

Prime Factorization: Write the prime factorization of each number in exponential form. Write the
prime factorization of the GCF by listing each prime factor that is common in the given numbers
raised to the smallest exponent. Multiply the factors to find the GCF.
V. Zabrocki 2011
page 1
M098
Carson Elementary and Intermediate Algebra 3e
Example 2:
Section 6.1
Find the GCF for 36 and 80.
Listing Method
36: 1, 2, 3, 4, 6, 9, 12, 18, 36
80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80
GCF = 4
Prime Factorization Method
2
2
36 = 2 · 3
4
80 = 2 · 5
2
GCF = 2 = 4
Generally, to find the GCF for monomials, find the GCF for the coefficients. Then write the variables that
are common to each term and use the smallest exponent.
6
4 2
3 5
Example 3: Find the GCF for 20m n, 10m n , 35m n
GCF for 20, 10 and 35 is 5.
The smallest exponent for m is 3.
The smallest exponent for n is 1.
3
GCF = 5m n
3. Write a polynomial as a product of a monomial GCF and a polynomial.
Multiplication and factoring are opposite operations:
In multiplication, we know the factors and want to find the product.
2 · 5 = 10
In factoring, we know the product and want to find the factors.
10 = 2 · 5
To find the GCF of a polynomial, find the GCF for each of its terms. Divide each term of the polynomial
by the GCF. Factored form will be the product of the GCF and the result of the division.
Example 4: Write in factored form. 8y - 24
GCF is 8.
8y  24
8y 24


 y3
8
8
8
8y – 24 = 8(y - 3)
Check by distributing to be sure your factors are correct.
2 3
2
Example 5: Write in factored form. 12x y – 6xy
2
GCF is 6xy .
12x 2 y 3  6xy 2
6xy 2
2 3
2

12x 2 y 3
6xy 2

6xy 2
6xy 2
 2xy  1
2
12x y – 6xy = 6xy (2xy – 1)
Check by distributing to be sure your factors are correct.
V. Zabrocki 2011
page 2
M098
Carson Elementary and Intermediate Algebra 3e
Section 6.1
When the first term is negative, factor out the negative along with the GCF. That changes the sign of
EVERY term left in the parentheses.
2
Example 6: Write in factored form. – 6a c – 18abc + 12ac
2
GCF is 6ac. Because the first term is negative, factor out the negative of the GCF. -6ac
 6a 2c  18abc  12ac 2
 6a 2c 18abc 12ac 2



 a  3b  2c
 6ac
 6ac
 6ac
 6ac
2
2
-6a c – 18abc + 12ac = -6ac(a + 3b – 2c)
Check by distributing to be sure your factors are correct.
Binomial factors can also be part of the GCF. Notice that x(x – 5) + 8(x – 5) is NOT in factored form
because it is the sum of two terms: x(x – 5) and 8(x – 5). (x – 5) is a factor in both terms, so it can be
factored out and the polynomial can be written in factored form.
x(x – 5) + 8(x – 5) = (x – 5)(x + 8)
Example 7: Write in factored form. 4m(2x – 3y) – 5(2x – 3y)
GCF is (2x – 3y).
4m(2x - 3y) - 5(2x - 3y)
4m(2x  3y) 5(2x  3y)


 4m  5
( 2x  3 y )
( 2x  3 y )
( 2x  3 y )
4m(2x – 3y) – 5(2x – 3y) = (2x – 3y)(4m – 5)
Place each binomial in parentheses.
4. Factor by grouping.
If the polynomial has 4 terms, it may be possible to factor by grouping. Always look for a common factor
before proceeding.
1. Group the first two terms and the last two terms. If the middle sign is negative, rewrite as an
addition.
2. Factor each binomial. If there are no common factors, switch the middle two terms and try
again.
3. If there is a common binomial, rewrite the expression in factored form.
Example 8: Factor by grouping.
2ab + 8a + 3b + 12
(2a + 8a) + (3b + 12)
Group the first two terms and the last two terms.
2a(b + 4) + 3(b + 4)
Factor each binomial. Distribute each one back to be sure
no errors have been made.
If there is a common binomial, rewrite the expression in
factored form.
(b + 4)(2a + 3)
Because multiplication is commutative, this could also be written (2a + 3) (b + 4).
V. Zabrocki 2011
page 3
M098
Carson Elementary and Intermediate Algebra 3e
4
4
3
4
3
y + 4y – by – 4b
Example 9: Factor by grouping.
y + 4y + – by – 4b
Section 6.1
If the middle sign is negative, rewrite as an addition.
3
(y + 4y ) + (– by – 4b)
Group the first two terms and the last two terms.
3
y (y + 4) + -b(y + 4)
Factor each binomial. Factor out the negative of the GCF if
the first term is negative. Change all signs in the
parentheses. Distribute each one back to be sure no errors
have been made.
If there is a common binomial, rewrite the expression in
factored form.
3
(y + 4)(y – b)
3
Because multiplication is commutative, this could also be written (y – b) (y + 4).
Example 10: Factor by grouping.
2
2
2
2
(m p – 3n )+ (n p – 3m )
2
2
2
2
m p + n p – 3n – 3m
2
2
2
2
(m p + n p) + ( – 3n – 3m )
2
2
2
2
p(m + n ) + -3(n + m )
2
2
(n + m )(p – 3)
2
2
2
Group the first two terms and the last two terms.
Factor each binomial. If there are no common factors,
switch the middle two terms and try again. Keep the
sign with each term.
If there is a common binomial, rewrite the expression in
factored form.
Example 11: Factor by grouping.
bx – by – ax - ay
(bx - by)+ (-ax - ay)
Group the first two terms and the last two terms.
b(x – y) + – a(x + y)
Factor each binomial
Prime
There is no common factor so this polynomial cannot be
factored.
Example 12: Factor completely.
2
2
2
5b(2a b – 2b + 3a – 3b)
2
2
2
5b( [2a b – 2b ]+ [3a – 3b] )
2
2
5b( 2b[a – b] + 3[a – b] )
2
5b(a – b)(2b + 3)
V. Zabrocki 2011
2
m p – 3n + n p – 3m
2 2
3
2
10a b – 10b + 15a b – 15b
2
Factor out the GCF.
Group the first two terms and the last two terms.
Factor each binomial.
If there is a common binomial, rewrite the expression in
factored form.
page 4