Sec 5.1 – Greatest Common Factor
Recall that factoring is the opposite of
multiplying.
Multiplying : 2 × 3 × 5 = 30
Factoring : 30 = 2 × 3 × 5
Sec 5.1 – Greatest Common Factor
Definition
The greatest common factor of a list of integers
is the largest common factor of those integers.
Example
3 is a common factor of 30 and 42. But their
greatest common factor is 6.
Sec 5.1 – Greatest Common Factor
Number
Divisible by
Must have the following property
2
End in 0, 2, 4, 6, or, 8
3
Sum of digits divisible by 3
4
Last two digits of number are divisible
by 4
5
Ends in 0 or 5
Sec 5.1 – Factoring Hints
Number
Divisible by
Must have the following property
6
Divisible by both 2 and 3
8
Last three digits form a number
divisible by 8
9
Sum of its digits is divisible by 9
10
Ends in 0
Sec 5.1 – Factoring Hints
Examples
124: Since 24 is divisible by 4, we know 124 is
divisible by 4. 124 = 31 x 4
125: Ends in 5, so it must be divisible by 5.
125 = 25 x 5.
126: The sum of the digits of 126 equals 9, so it
must be divisible by 9. 126 = 14 x 9.
Sec 5.1 – Greatest Common Factor
1.
2.
3.
4.
Factor – Write each number in prime factored
form.
List common factors - List each prime number
that is a factor of every number in the list.
Choose smallest exponents – Use as exponents
on the common prime factors the smallest
exponent from the prime factored forms.
Multiply – Multiply the primes from Step 3. If
there are no common primes, the greatest
common factor is 1.
Sec 5.1 – Greatest Common Factor
Example
Find the greatest common factor of 120
and 252
Sec 5.1 – Greatest Common Factor
1. Factor
120 = 12 × 10
= 4 × 3× 2 × 5
3
= 2 × 3× 5
252 = 9 × 28
= 9× 4× 7
2
2
= 3 ×2 ×7
Sec 5.1 – Greatest Common Factor
1. List Common Factors
120 = 2 × 3 × 5
2
2
252 = 3 × 2 × 7
3
Common Factors
2, 3
Sec 5.1 – Greatest Common Factor
1. Choose smallest exponents
120 = 2 × 3 × 5
2
2
252 = 3 × 2 × 7
3
Common Factors
2, 3
2
2 ,3
Sec 5.1 – Greatest Common Factor
1. Multiply
GCF = 2 2 × 3
= 12
120 = 12 × 10
252 = 12 × 21
Sec 5.1 – Greatest Common Factor
Example
Find the greatest common factor of 24
and 25
Sec 5.1 – Greatest Common Factor
1. Factor
24 = 8 × 3
3
= 2 ×3
2
25 = 5
Sec 5.1 – Greatest Common Factor
1. List Common Factors
24 = 2 × 3
2
25 = 5
3
Common Factors
1
GCF = 1
Sec 5.1 – Greatest Common Factor
We can also find the greatest common factor for a
list of variable terms.
Find the greatest common factor of:
5
7
12r ,15r , and 18r
9
Sec 5.1 – Greatest Common Factor
1. Factor
12r =
=
7
15r =
9
18r =
=
5
4 × 3× r
2
5
2 × 3× r
7
3× 5× r
9
2× 9× r
2
9
2× 3 × r
5
Sec 5.1 – Greatest Common Factor
1. List Common Factors
12r = 2 × 3 × r
7
7
15r = 3 × 5 × r
9
2
9
18r = 2 × 3 × r
2
2
Common Factors
3, r
5
Sec 5.1 – Greatest Common Factor
1. Choose smallest exponents
12r = 2 × 3 × r
7
7
15r = 3 × 5 × r
9
2
9
18r = 2 × 3 × r
2
2
Common Factors
3, r
Smallest Exponents
3, r
5
5
Sec 5.1 – Greatest Common Factor
1. Multiply
Smallest Exponents
3, r
5
GCF = 3r
5
Sec 5.1 – Factoring out GCF
We use the technique of finding the GCF to
write a polynomial in factored form.
Sec 5.1 – Factoring out GCF
Example
Write the polynomial expression 7x+35 in
factored form.
7 x + 35 = 7 x + 7 ⋅ 5
= 7( x + 5)
Sec 5.1 – Factoring out GCF
Factor:
8 p q + 16 p q + 12 p q
5
2
6
3
4
7
= 2 p q + 2 p q + 2 × 3p q
3
5
2
4
6
3
2
4
= 2 p q (2 p + 2 p q + 3q )
2
4
2
2
2
5
7
Sec 5.1 – Factoring by Grouping
2mn − 8n + 3m − 12
= 2n(m − 4) + 3(m − 4)
= (2n + 3)(m − 4)
Sec 5.1 – Factoring by Grouping
1. Group terms. Collect the terms into two groups so
that each group has a common factor
2. Factor within group. Factor out the greatest
common factor from each group.
3. Factor the entire polynomial. Factor a common
binomial factor from the results of
Step 2.
4. If necessary, rearrange terms. If Step 2 does not
result in a common binomial factor, try a different
grouping.
Sec 5.1 – Factoring by Grouping
Example
Factor
6 y − 20w + 15 yw − 8 yw
2
2
Sec 5.1 – Factoring by Grouping
1. Group terms. Collect the terms into two groups so
that each group has a common factor
6 y − 20w + 15 yw − 8 yw
2
2
= [6 y − 20 w ] + [15 yw − 8 yw]
2
2
Sec 5.1 – Factoring by Grouping
1. Factor within group. Factor out the greatest
common factor from each group.
6 y − 20w + 15 yw − 8 yw
2
2
= 2( y − w ) + 7 yw
2
2
1. If necessary, rearrange terms. If Step 2
does not result in a common binomial factor,
try a different grouping
Sec 5.1 – Factoring by Grouping
1. Group terms. Collect the terms into two groups so
that each group has a common factor
6 y − 20w + 15 yw − 8 yw
2
2
= [6 y + 15 yw] − [20 w + 8 yw]
2
2
Sec 5.1 – Factoring by Grouping
1. Factor within group. Factor out the greatest
common factor from each group.
[6 y + 15 yw] − [20w + 8 yw]
2
2
= 3 y (2 y + 5w) − 4w(5w + 2 y )
= 3 y (2 y + 5w) − 4w(2 y + 5w)
Sec 5.1 – Factoring by Grouping
1. Factor the entire polynomial. Factor a common
binomial factor from the results of
Step 2.
3 y (2 y + 5w) − 4w(2 y + 5w)
= (3 y − 4w)(2 y + 5w)
Sec 5.2 - Factoring Trinomials
Using FOIL, we have learned to expand the
product of two polynomials:
( x + 8)( x − 2) = x 2 − 2 x + 8 x − 16
= x 2 + 6 x − 16
Sec 5.2 - Factoring Trinomials
Now suppose we are given the expanded
polynomial and we want to write it in factored
form.
x 2 + 6 x − 16 = ( x + r1 )( x + r2 )
Sec 5.2 - Factoring Trinomials
We need to determine the values of r1 and r2
such that:
x 2 + 6 x − 16 = ( x + r1 )( x + r2 )
= x 2 + r1 x + r2 x + r1r2
Therefore,
= x 2 + (r1 + r2 ) x + r1r2
r1 + r2 = 6
r1r2 = − 16
Sec 5.2 - Factoring Trinomials
Observation
Since the product is negative and the sum
is positive, one constant must be positive
and the other negative.
Sec 5.2 - Factoring Trinomials
Factor x + 6 x − 16
2
Product
Sum
-1x16=-16
16+(-1)=15
-2x8=-16
8+(-2)=6
-4x4=16
4+(-4)=0
x 2 + 6 x − 16 = ( x + 8)( x + (− 2) ) = ( x + 8)( x − 2)
Sec 5.2 - Factoring Trinomials
Example
Factor the polynomial
x 2 + 12 x + 36
Observation
Since the product is positive and the
sum is positive, both constants must be
positive.
Sec 5.2 - Factoring Trinomials
Polynomial:
x 2 + 12 x + 36
Product
Sum
1x36=36
1+36=37
2x18=36
2+18=20
3x12=36
3+12=15
4x9=36
4+9=13
6x6=36
6+6=12
x 2 + 12 x + 36 = ( x + 6 )( x + 6 )
Sec 5.2 - Factoring Trinomials
Example
Factor the polynomial
x 2 − 11x + 18
Observation
Since the product is positive and the
sum is negative, both constants must be
negative.
Sec 5.2 - Factoring Trinomials
Polynomial: x 2 − 11x + 18
Product
Sum
(-1)x(-18)=18 -1+(-18)=-19
(-2)x(-9)=18
-2+(-9)=-11
(-3)x(-6)=18
-3+(-6)=-9
x 2 − 11x + 18 = ( x + ( − 2) ) ( x + ( − 9 ) ) = ( x − 2 )( x − 9 )
Sec 5.2 - Factoring Trinomials
Example
2
x
+ 9 x + 21
Factor the polynomial
Product
Sum
1x21=21
1+21=22
3x7=21
3+7=10
Sec 5.2 - Factoring Trinomials
x2+9x+21 is called a prime polynomial
since it cannot be factored using only
integer factors.
Sec 5.2 - Factoring Trinomials
Polynomial:
r 2 − 6rs + 8s 2
Product
Sum
(-1s)x(-8s)=8s2
-1s+(-8s)=-9s
(-2s)x(-4s)=8s2
-2s+(-4s)=-6s
r 2 − 6rs + 8s 2 = ( r + ( − 2s ) ) ( r + ( − 4s ) ) = ( r − 2s )( r − 4s )
Sec 5.2 - Factoring Trinomials with a
Common Factor
Important
Before trying to factor a polynomial, first
factor out the greatest common factor.
Sec 5.2 - Factoring Trinomials with a
Common Factor
2
2
3x 4 − 15 x 3 + 18 x 2 = 3x ( x − 5 x + 6)
= 3x 2 ( x − 2)( x − 3)
Sec 5.3 – Coefficient of Square Term
not 1
When we factored a trinomial like x2+12x+36,
we looked for two numbers r1 and r2 whose sum
were 12 and their product were 36 so that we
could write the polynomial as (x+r1)(x+r2).
Sec 5.3 - Coefficient of Square Term
not 1
If the coefficient of the x2 term is not one, we
use a different technique. Instead of directly
factoring the polynomial, we use factor by
grouping as in Section 5.1.
Sec 5.3 - Coefficient of Square Term
not 1
Problem
Factor 10x2-23x+12 by grouping
10 x 2 − 23x + 12 = 10 x 2 − 8 x − 15 x + 12
= 2 x( 5 x − 4 ) − 3( 5 x − 4 )
= ( 2 x − 3)( 5 x − 4 )
Sec 5.3 - Coefficient of Square Term
not 1
How do we know to break up -23x into -8x - 15x?
( 2 x − 3)( 5 x − 4)
= 2 ⋅ 5 x 2 + 2 ⋅ ( − 4) x + ( − 3) ⋅ 5 x + ( − 3) ⋅ ( − 4)
− 8 = 2 ⋅ ( − 4) = c1
− 15 = ( − 3) ⋅ 5 = c2
− 8 + ( − 15) = c1 + c2 = − 23
Clearly, the sum of -8 and -15 must be -23.
Sec 5.3 - Coefficient of Square Term
not 1
Notice also that
c1 ⋅ c2 = − 8 ⋅ ( − 15)
= 2 ⋅ ( − 4 ) ⋅ ( − 3) ⋅ 5
= 2 ⋅ 5 ⋅ ( − 4 ) ⋅ ( − 3)
= 10 ⋅ 12
which is the product of the first and last constants.
Sec 5.3 - Coefficient of Square Term
not 1
Therefore, to factor
10 x 2 − 23x + 12
by grouping, we look for 2 numbers c1 and c2
such that
c1 + c2 = − 23
c1 × c2 = 10 ⋅ 12 = 120
Sec 5.3 - Coefficient of Square Term
not 1
Factor 6p2+19p+10
c1 + c2 = 19
c1 × c2 = 60
Since the sum and product are positive, both
constants must be positive.
Sec 5.3 - Factoring Trinomials with
All Terms Positive
Factor 6p2+19p+10:
Product
Sum
1x60=60
1+60=61
2x30=60
2+30=32
3x20=60
3+20=23
4x15=60
4+15=19
Sec 5.3 - Factoring Trinomials with
All Terms Positive
So we have:
6 p + 19 p + 10 = 6 p + 4 p + 15 p + 10
= 2 p(3 p + 2) + 5(3 p + 2)
= (2 p + 5)(3 p + 2)
2
2
Sec 5.3 - Factoring Trinomials with a
Negative Middle Term
Factor 10m2-23m+12
c1 + c2 = − 23
c1 × c2 = 120
Since the sum is negative and the product is
positive, both constants must be negative.
Sec 5.3 - Factoring Trinomials with a
Negative Middle Term
Factor 10m2-23m+12:
Product
Sum
-1x-120=120
-1+(-120)=121
-2x-60=120
-2+(-60)=-62
-3x-40=120
-3+(-40)=-43
-4x-30=120
-4+(-30)=-34
Sec 5.3 - Factoring Trinomials with a
Negative Middle Term
Factor 10m2-23m+12:
Product
Sum
-5x-24=120
-5+(-24)=-29
-6x-20=120
-6+(-20)=-26
-8x-15=120
-8+(-15)=-23
Sec 5.3 - Factoring Trinomials with a
Negative Middle Term
So we have:
10m − 23m + 12 = 10m − 15m − 8m + 12
= 5m(2m − 3) − 4(2m − 3)
= (5m − 4)(2m − 3)
2
2
Sec 5.3 - Factoring Trinomials with a
Negative Last Term
Factor 5x2+13x-6
c1 + c2 = 13
c1 × c2 = − 30
Since the sum is positive and the product is
negative, one constant must be positive and the
other negative.
Sec 5.3 - Factoring Trinomials with a
Negative Last Term
Factor 5x2+13x-6:
Product
Sum
-1x30=-30
30+(-1)=29
-2x15=-30
15+(-2)=13
Sec 5.3 - Factoring Trinomials with a
Negative Last Term
So we have:
5 x + 13x − 6 = 5 x + 15 x − 2 x − 6
= 5 x( x + 3) − 2( x + 3)
= (5 x − 2)( x + 3)
2
2
Sec 5.3 - Factoring Trinomials with
Two Variables
Factor 6x2+11xy-10y2
c1 + c2 = 11y
c1 × c2 = − 60 y
2
Since the sum is positive and the product is
negative, one constant must be positive and the
other negative.
Sec 5.3 - Factoring Trinomials with
Two Variables
Factor 6x2+11xy-10y2 :
Product
Sum
− 1 y × 60 y = − 60 y 2
60 y + ( − 1y ) = 59 y
− 2 y × 30 y = − 60 y 2
30 y + ( − 2 y ) = 28 y
− 3 y × 20 y = − 60 y 2
20 y + ( − 3 y ) = 17 y
− 4 y × 15 y = − 60 y 2
15 y + ( − 4 y ) = 11y
Sec 5.3 - Factoring Trinomials with
Two Variables
So we have:
6 x + 11xy − 10 y = 6 x − 4 xy + 15 xy − 10 y
2
2
2
2
= 2 x(3x − 2 y ) + 5 y (3x − 2 y )
= (2 x + 5 y )(3x − 2 y )
Sec 5.3 - Factoring Trinomials with a
Common Factor
Factor 15x3+55x2+30x
First factor out the GCF :
15 x + 55 x + 30 x
3
2
= 5 x (3x + 11x + 6)
2
Sec 5.3 - Factoring Trinomials with a
Common Factor
Since we now have 5x(3x2+11x+6):
c1 + c2 = 11
c1 × c2 = 18
Since the sum and product is positive, each
constant must be positive.
Sec 5.3 - Factoring Trinomials with a
Common Factor
Factor 5x(3x2+11x+6):
Product
Sum
1x18=18
1+18=19
2x9=18
2+9=11
Sec 5.3 - Factoring Trinomials with a
Common Factor
So we have:
3
15 x + 55 x + 30 x = 5 x(3x + 11x + 6)
2
= 5 x(3x + 2 x + 9 x + 6)
= 5 x[ x(3x + 2) + 3(3x + 2)]
= 5 x( x + 3)(3x + 2)
2
2
Sec 5.3 - Factoring Trinomials with a
Negative Initial Term
Observation
If the coefficient of the highest order term
is negative, it may be easier to factor the
polynomial by initially factoring a -1 from
each term.
− 4 x + 2 x + 30 = − 2( 2 x − x − 15)
2
2
Sec 5.4 - Factoring the Difference of
Two Squares
We have previously derived the formula
(x+y)(x-y)=x2-y2
We can turn this around to derive a formula for
factoring a polynomial:
x2-y2=(x+y)(x-y)
Sec 5.4 - Factoring the Difference of
Two Squares
Example
Factor x2-100
x 2 − 100 = x 2 − 10 2
= ( x + 10)( x − 10)
Sec 5.4 - Factoring the Difference of
Two Squares
Example
Factor x2+100
This is a prime polynomial since it is not the
difference of two squares.
Note:
( x + 10)( x + 10) = x + 20 x + 100
2
( x − 10)( x − 10) = x − 20 x + 100
2
Sec 5.4 - Factoring the Difference of
Two Squares
Example
Factor 50r2-32
50r 2 − 32 = 2(25r 2 − 16)
= 2[(5r ) 2 − 4 2 ]
= 2(5r + 4)(5r − 4)
Sec 5.4 - Factoring a Perfect Square
Trinomial
Example
Factor x2+6x+9
Using our previous methods we would look for
two constants a and b such that:
r1 + r2 = 6
r1 × r2 = 9
Sec 5.4 - Factoring a Perfect Square
Trinomial
Factor x2+6x+9
Product
Sum
1x9=9
1+9=10
3x3=9
3+3=6
Sec 5.4 - Factoring a Perfect Square
Trinomial
Factor x2+6x+9
x 2 + 6 x + 9 = ( x + 3)( x + 3)
= ( x + 3) 2
Sec 5.4 - Factoring a Perfect Square
Trinomial
Factor x2+6x+9
Note
This is a special case of our method in section
5.2, where r1=r2.
r1 + r2 = 2r1 = 6
2
r1 × r2 = r1 = 9
Sec 5.4 - Factoring a Perfect Square
Trinomial
Factor x2+6x+9
 You should first check to see if the constant
term is a perfect square, in this case 9.
 Then, if the middle term is twice this amount
(2x3=6), you have one of the special cases:
x 2 + 2ax + a 2 = ( x + a ) 2
x 2 − 2ax + a 2 = ( x − a ) 2
Sec 5.4 - Factor the Difference of
Two Cubes
To factor the difference of two cubes, we use
the formula:
x − y = ( x − y )( x + xy + y )
3
3
2
2
Sec 5.4 - Factor the Difference of
Two Cubes
Example
Factor 8z3-125
Since, 8z3=(2z)3 and 125=53, let x=2z and
y=5
(
8 z 3 − 125 = ( 2 z − 5) ( 2 z ) + ( 2 z )( 5) + ( 5)
2
= ( 2 z − 5) ( 4 z 2 + 10 z + 25)
2
)
Sec 5.4 - Factor the Sum of Two
Cubes
To factor the sum of two cubes, we use the
formula: 3 3
2
2
x + y = ( x + y )( x − xy + y )
Sec 5.4 - Factor the Sum of Two
Cubes
Example
Factor 27z3+1000
Since, 27z3=(3z)3 and 1000=103, let x=3z
and y=10
(
27 z 3 + 1000 = (3z + 10) ( 3 z ) − (10 z ) ⋅ 3 + (10)
2
= (3z + 10)(9 z 2 − 30 z + 100)
2
)
Sec 5.5 - Solving Quadratic
Equations by Factoring
Definition
An equation of the form
ax2+bx+c=0
where a, b, and c are real numbers, and a is not
equal to zero, is called a quadratic equation.
Examples:
2 x 2 + 11x + 14 = 0
x2 + 6x = 0
x2= 9
Sec 5.5 - Zero-Factor Property
Zero-Factor Property
If a and b are real numbers, and if ab=0, then at
least one of the numbers a or b must be equal to
zero.
Sec 5.5 - Using the Zero-Factor
Property
Example
Solve (5x+7)(2x+3)=0
5x + 7 = 0
or
7
x= −
5
2x + 3 = 0
3
x= −
2
Sec 5.5 - Solving a Quadratic Not in
Standard Form
Example
Solve x2=8-2x2
x + 2x − 8 = 0
( x + 4)( x − 2) = 0
x + 4 = 0 or
x= −4
x− 2= 0
x= 2
Sec 5.5 - Solving a Quadratic
Equation by Factoring
1.
2.
3.
4.
Write in standard form. Place all terms on one side of
the equals sign with zero on the other.
Factor. Factor the equation completely.
Use the zero-factor property. Set each factor with a
variable equal to zero, and solve the resulting
equation.
Check. Check each solution in the original equation.
Sec 5.5 - Solving a Quadratic with a
Common Factor
Example
Solve 3z2-9z=30
3 z 2 − 9 z − 30 = 0
3( z 2 − 3 z − 10) = 0
3( z − 5)( z + 2) = 0
z− 5= 0
z= 5
or
z+ 2= 0
z = −2
Note: The common constant factor has nothing to
do with the solution
Sec 5.5 - Solving a Quadratic with a
Common Factor
Example
Solve 3z2=9z
3z 2 − 9 z = 0
3 z ( z − 3) = 0
3z = 0
z= 0
or
z− 3= 0
z= 3
Note: The common factor 3z is include in the
solution because it contains a variable.
Also, don’t divide each side by 3z. You
would lose the solution z=0.
Sec 5.5 - Solving Special Quadratic
Equations
Example
(2x+1)(2x2+7x-15)=0
(2 x + 1)(2 x 2 + 7 x − 15) = 0
(2 x + 1)(2 x 2 − 3 x + 10 x − 15) = 0
( 2 x + 1)[ x ( 2 x − 3) + 5( 2 x − 3)] = 0
( 2 x + 1)( x + 5)(2 x − 3) = 0
2x + 1 = 0
1
x= −
2
or
x+ 5= 0
x = −5
or
2x − 3 = 0
3
x=
2
Sec 5.5 - Equations with More than
two Variable Factors
Example
Solve 49x2=9
49 x 2 − 9 = 0
(7 x − 3)(7 x + 3) = 0
7x − 3 = 0
3
x=
7
or
7x + 3 = 0
3
x= −
7
Sec 5.6 - Applications of Quadratic
Equations
Example: Area Problem
In a home, the length of a hall is five times the
width. The area of the floor is 45 square feet.
Find the length and width of the hall.
Sec 5.6 - Solving an Area Problem
Solution:
Let x = width of the hall
5x = length of the hall
A=
45 =
0=
0=
l× w
x(5 x)
5 x 2 − 45
5( x 2 − 9)
0 = 5( x − 3)( x + 3)
x=3 or x=-3
Sec 5.6 - Applications of Quadratic
Equations
Example:
Area and Perimeter Problem
The width of a rectangular yard is 4 meters less
than the length. The area is numerically 92 more
than the perimeter. Find the length and width of
the yard.
Sec 5.6 - Solving an Area and
Perimeter Problem
Solution:
Let x = length of the yard
x - 4 = width of the yard
P = 2(l + w)
A= l× w
x( x − 4) = 2( x + x − 4) + 92
x 2 − 4 x = 4 x + 84
0 = x 2 − 8 x − 84
0 = ( x − 14)( x + 6)
x=14 or x=-6
Sec 5.6 - Solving Problems Using the
Pythagorean Formula
Pythagorean Formula
If a right triangle (a triangle with a 90° angle)
has longest side of length c and two sides of
length a and b, then
a2+b2=c2
c
a
b
Sec 5.6 - Applications of Quadratic
Equations
Example: Pythagorean Theorem
The hypotenuse (side opposite the right angle) of a right
triangle is three inches longer than the longer leg. The
shorter leg is three inches shorter than the longer leg. Find
the lengths of the sides of the triangle.
x+3
x-3
x
Sec 5.6 - Using the Pythagorean
Formula
Solution:
( x + 3) 2 = x 2 + ( x − 3) 2
x2+ 6x + 9 = x2 + x2 − 6x + 9
0 = x 2 − 12 x
= x ( x − 12)
12 inches, 9 inches, and 15 inches
Sec 5.6 - Applications of Quadratic
Equations
Example: Rocket Problem
If an object is propelled upward from ground level
with an initial velocity 64ft/sec, its height h in feet t
seconds later is given by the equation
h=-16t2+64t.
At what time will it be 48 feet above the ground?
Sec 5.6 - Rocket Problem
Solution:
h = − 16t 2 + 64t
48 = − 16t 2 + 64t
0 = 16t 2 + 64t − 48
0 = − 16(t 2 − 4t + 3)
0 = − 16(t − 3)(t − 1)
t=1 or t=3