Transcript Factor out the GCF.

Section 5.1
Quadratic Equations
OBJECTIVES
A
Find the greatest
common factor (GCF) of
numbers.
B
Find the GCF of terms.
OBJECTIVES
C
Factor out the GCF.
D
Factor a four-term
expression by
grouping.
DEFINITION
Greatest Common Factor
(GCF)
The largest common factor of
the integers in a list.
PROCEDURE
Finding the Product
4(x + y) = 4x + 4y
5(a – 2b) = 5a – 10b
2
2x(x + 3) = 2x + 6x
PROCEDURE
Finding the Factors
4x + 4y = 4(x + y)
5a – 10b = 5(a – 2b)
2
2x + 6x = 2x(x + 3)
DEFINITION
GCF of a Polynomial
The term ax n is the GCF
of a polynomial if
1. a is the greatest integer that
divides each coefficient.
DEFINITION
GCF of a Polynomial
The term ax n is the GCF
of a polynomial if
2. n is the smallest exponent of
x in all the terms.
Chapter 5
Factoring
Section 5.1
Exercise #2
Find the GCF of 18x 2 y 4 and 30x 3 y 5 .
2 18
30
3
9
15
3
5
GCF = 2 • 3 = 6
GCF = 6x 2 y 4
x 2y 4
x3y5
x 2y 4
Chapter 5
Factoring
Section 5.1
Exercise #5
Factor 2x 3 + 6x 2 y + x + 3y.
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 2x 3 + 6x 2 y + x + 3y.


= 2x 3 + 6x 2 y +  x + 3y 
= 2x 2  x + 3y  + 1 x + 3y 


=  x + 3y  2x 2 + 1
Section 5.2
Quadratic Equations
OBJECTIVES
A
Factor trinomials of the
2
form x + bx + c.
RULE
Factoring Rule 1 (F1)
2
X + (A + B)X + AB = (X + A)(X + B)
PROCEDURE
2
Factoring x + bx + c
Find two integers whose product
is c and whose sum is b.
1. If b and c are positive, both
integers must be positive.
PROCEDURE
2
Factoring x + bx + c
Find two integers whose product
is c and whose sum is b.
2. If c is positive and b is
negative, both integers
must be negative.
PROCEDURE
2
Factoring x + bx + c
Find two integers whose product
is c and whose sum is b.
3. If c is negative, one integer
must be negative and one
positive.
Chapter 5
Factoring
Section 5.2
Exercise #6
Factor x 2 – 8x + 12.
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor x 2 – 8x + 12.
=  x – 6  x – 2 
Section 5.3
Quadratic Equations
OBJECTIVES
A
Use the ac test to
determine whether
2
ax + bx + c is factorable.
OBJECTIVES
B
2
Factor ax + bx + c
by grouping.
OBJECTIVES
C
2
Factor ax
+ bx + c
using FOIL.
TEST
2
ac test for ax + bx + c
A trinomial of the form
2
ax + bx + c is factorable if
there are two integers with
product ac and sum b.
TEST
ac test
We need two numbers
whose product is ac.
ax 2 + bx + c
The sum of the numbers
must be b.
PROCEDURE
Factoring by FOIL
Product must be c.
ax 2 + bx + c = (__x + __)(__x + __)
Product must be a.
PROCEDURE
Factoring by FOIL
1. The product of the numbers in
the first (F) blanks must be a.
PROCEDURE
Factoring by FOIL
2. The coefficients of the
outside (O) products and the
inside (I) products must add
up to b.
PROCEDURE
Factoring by FOIL
3. The product of numbers in
the last (L) blanks must be c.
Chapter 5
Factoring
Section 5.3
Exercise #8
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 6x 2 – 11xy + 3y 2 .
 3x
– y  2 x – 3y

Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Section 5.4
Quadratic Equations
OBJECTIVES
A
Recognize the square of
a binomial (a perfect
square trinomial).
OBJECTIVES
B
Factor a perfect
square trinomial.
OBJECTIVES
C
Factor the difference
of two squares.
RULES
Factoring Rules 2 and 3:
PERFECT SQUARE
TRINOMIALS
2
X
+ 2AX
2
+A
= (X
2
+ A)
(F2)
Note that X 2+ A2  (X + A)2
RULES
Factoring Rules 2 and 3:
PERFECT SQUARE
TRINOMIALS
X2
– 2AX
Note that
2
+A
X2
–
= (X – A)2 (F3)
2
A
 (X – A)2
RULE
Factoring Rule 4:
THE DIFFERENCE OF TWO
SQUARES
2
X –
2
A = (X + A)(X – A) (F4)
Chapter 5
Factoring
Section 5.4
Exercise #11
Factor 9x 2 – 12xy + 4y 2 .
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 9x 2 – 12xy + 4y 2 .

= 3x – 2y

2
Chapter 5
Factoring
Section 5.4
Exercise #13
Factor 16x 2 – 25y 2 .
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 16x 2 – 25y 2 .
=  4 x + 5y  4 x – 5 y 
Section 5.5
Quadratic Equations
OBJECTIVES
A
Factor the sum or
difference of two cubes.
OBJECTIVES
B
Factor a polynomial by
using the general
factoring strategy.
OBJECTIVES
C
Factor expressions
whose leading
coefficient is –1.
RULE
Factoring Rule 5:
THE SUM OF TWO CUBES.
(F5)
X 3+ A3 = (X + A)(X 2 – AX + A2)
RULE
Factoring Rule 6:
THE DIFFERENCE OF TWO
(F6)
CUBES.
X 3 – A3 = (X – A)(X 2+ AX + A2)
PROCEDURE
General Factoring Strategy
1. Factor out all common
factors.
PROCEDURE
General Factoring Strategy
2. Look at the number of
terms inside the
parentheses. If there are:
Four terms:
Factor by grouping.
PROCEDURE
General Factoring Strategy
Three terms:
If the expression is a perfect
square trinomial, factor it.
Otherwise, use the ac test to
factor.
PROCEDURE
General Factoring Strategy
Two terms and squared:
Look at the difference of two
2
2
squares (X –A ) and factor it.
2
2
Note: X +A is not factorable.
PROCEDURE
General Factoring Strategy
Two terms and cubed:
Look for the sum of two cubes
3
3
(X +A ) or the difference of two
3 3
cubes (X -A ) and factor it.
PROCEDURE
General Factoring Strategy
Make sure your expression is
completely factored.
Check by multiplying the
factors you obtain.
Chapter 5
Factoring
Section 5.5
Chapter 5
Factoring
Section 5.5
Exercise #15
Factor 8y 3 – 125x 3 .
3
=  2y  –  5 x 
3
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 8y 3 – 125x 3 .
  – 5x 
= 2y

3
3

= 2y – 5x 4y 2 + 10xy + 25x 2

Chapter 5
Factoring
Section 5.5
Exercise #17
Factor 2x 3 – 8x 2 – 10x.
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 2x 3 – 8x 2 – 10x.

= 2x x 2 – 4x – 5

= 2 x  x – 5  x + 1
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor 2x 3 + 6x 2 + x + 3.

 
= 2 x 3 + 6x 2 + x + 3
= 2 x 2  x + 3  + 1 x + 3 


=  x + 3 2x 2 + 1

Chapter 5
Factoring
Section 5.5
Exercise #20
Factor – 9x 4 + 36x 2 .
Factoring Strategy Flow Chart
Factor out GCF
(2) Terms
(3) Terms
Difference of
Squares
Perfect Square
Trinomial
Sum/Difference
of Cubes
(x2 + bx + c)
(ax2 + bx + c)
(4) Terms
Grouping
Factor – 9x 4 + 36x 2 .

= – 9x 2 x 2 – 4

= – 9 x 2  x + 2  x – 2 
Section 5.6
Quadratic Equations
OBJECTIVES
A
Solve quadratic
equations by factoring.
DEFINITION
Quadratic Equation in
Standard Form
If a, b and c are real numbers
(a  0),
2
ax + bx
+c =0
PROCEDURE
Solving Quadratics by
Factoring
1. Perform necessary
operations on both sides so
that right side = 0.
PROCEDURE
Solving Quadratics by
Factoring
2. Use general factoring
strategy to factor the left
side if necessary.
PROCEDURE
Solving Quadratics by
Factoring
3. Use the principle of zero
product and make each
factor on the left equal 0.
PROCEDURE
Solving Quadratics by
Factoring
4. Solve each of the resulting
equations.
PROCEDURE
Solving Quadratics by
Factoring
5. Check results by
substituting solutions
obtained in step 4 in
original equation.
Chapter 5
Factoring
Section 5.6
Exercise #24
Solve. (2x – 3)(x – 4) = 2(x – 1) – 1
2x 2 – 8x – 3x + 12 = 2x – 2 – 1
2x 2 – 11x + 12 = 2x – 3
2x 2 – 13x + 12 = – 3
2x 2 – 13x + 15 = 0
(2x – 3)(x – 5) = 0
2x – 3 = 0 or
x –5 =0
Solve. (2x – 3)(x – 4) = 2(x – 1) – 1
2x – 3 = 0 or
2x = 3
3
x=
2
x –5 =0
x =5
Section 5.7
Quadratic Equations
OBJECTIVES
A
Integer problems.
B
Area and perimeter
problems.
OBJECTIVES
C
Problems involving the
Pythagorean Theorem.
D
Motion problems.
NOTE
Notation
Terminology
2 consecutive
integers
n, n+1
Examples: 3,4; –6,–5
NOTE
Notation
Terminology
3 consecutive
integers
n, n+1, n+2
Examples: 7, 8, 9; – 4,– 3,– 2
NOTE
Notation
Terminology
2 consecutive
even integers
n, n +2
Examples: 8,10; – 6,– 4
NOTE
Notation
Terminology
2 consecutive
odd integers
n, n +2
Examples: 13,15; – 21,–19
DEFINITION
Pythagorean Theorem
If the longest side of a
right triangle is of length c
and the other two sides
are of length a and b, then
a2 + b2 = c2
DEFINITION
Pythagorean Theorem
a 2 + b2 = c 2
Leg a
Hypotenuse
c
Leg b
Chapter 5
Factoring
Section 5.7
Exercise #26
The product of two consecutive odd integers is 13
more than 10 times the larger of the two integers.
Find the integers.
Let x = 1st odd integer
Let x + 2 = 2nd odd integer
x(x + 2) = 10(x + 2) + 13
x 2 + 2x = 10x + 20 + 13
x 2 + 2x = 10x + 33
x 2 – 8x = 33
x 2 – 8x – 33 = 0
The product of two consecutive odd integers is 13
more than 10 times the larger of the two integers.
Find the integers.
Let x = 1st odd integer
Let x + 2 = 2nd odd integer
x 2 – 8x – 33 = 0
(x – 11)(x + 3) = 0
x – 11 = 0
or
x = 11 or
x +3=0
x =–3
The product of two consecutive odd integers is 13
more than 10 times the larger of the two integers.
Find the integers.
Let x = 1st odd integer
Let x + 2 = 2nd odd integer
x = 11
x + 2 = 13
or
x =–3
x +2=–1
The integers are 11 and 13 or – 3 and – 1.
Chapter 5
Factoring
Section 5.7
Exercise #29
A rectangular 10-inch television screen (measured
diagonally) is 2 inches wider than it is high. What are
the dimensions of the screen?
Let x = height
Let x + 2 = length
10
x +2
x 2 + (x + 2) 2 = 102
x
x 2 + (x + 2) 2 = 102
x 2 + x 2 + 4x + 4 = 100
2x 2 + 4x + 4 = 100
2x 2 + 4x – 96 = 0
x 2 + 2x – 48 = 0
(x + 8)(x – 6) = 0
x + 8 = 0 or x – 6 = 0
x =–8
x =6
height  –8
x +2=8
The screen is 6 inches high and 8 inches long.