Section 5.1 Quadratic Equations. OBJECTIVES A Find the greatest common factor (GCF) of numbers. B Find the GCF of terms.
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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.
DEFINITIONGreatest Common Factor (GCF)
The largest common factor of the integers in a list.
PROCEDUREFinding the Product
4(x + y) = 4x + 4y
5(a – 2b) = 5a – 10b
2x(x + 3) = 2x2 + 6x
PROCEDUREFinding the Factors
4x + 4y = 4(x + y)
5a – 10b = 5(a – 2b)
2x2 + 6x = 2x(x + 3)
DEFINITIONGCF of a Polynomial
The term axn is the GCF of a polynomial if
1. a is the greatest integer that divides each coefficient.
DEFINITIONGCF of a Polynomial
The term axn is the GCF of a polynomial if
2. n is the smallest exponent of x in all the terms.
Section 5.1Exercise #2
Chapter 5Factoring
18 30 2
9 15 3
3 5
GCF = 2 • 3 = 6
Find the GCF of 18x2y4 and 30x3y5 .
x2y4
x3y5
x2y4
GCF = 6x2y4
Section 5.1Exercise #5
Chapter 5Factoring
Factor 2x3 + 6x2y + x + 3y.
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 2x3 + 6x2y + x + 3y.
3 2 = 2 + 6 + + 3x x y x y
2 = 2 + 3 + 1 + 3x x y x y
2 = + 3 2 + 1x y x
Section 5.2
Quadratic Equations
OBJECTIVES
A Factor trinomials of the form x 2 + bx + c.
RULEFactoring Rule 1
X 2 + (A + B)X + AB = (X + A)(X + B)
(F1)
PROCEDUREFactoring x2 + bx + cFind two integers whose product is c and whose sum is b.
1. If b and c are positive, both integers must be positive.
PROCEDUREFactoring x2 + bx + cFind 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.
PROCEDUREFactoring x2 + bx + cFind two integers whose product is c and whose sum is b.
3. If c is negative, one integer must be negative and one positive.
Section 5.2Exercise #6
Chapter 5Factoring
Factor x2 – 8x + 12.
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor x2 – 8x + 12.
= – 6 – 2x x
Section 5.3
Quadratic Equations
OBJECTIVES
A Use the ac test to determine whether ax 2 + bx + c is factorable.
OBJECTIVES
B
Factor ax2 + bx + c by grouping.
OBJECTIVES
C
Factor ax2 + bx + c using FOIL.
TEST
ac test for ax2 + bx + c
A trinomial of the form ax2 + bx + c is factorable if there are two integers with product ac and sum b.
TESTac test
ax2 + bx + c
We need two numbers whose product is ac.
The sum of the numbers must be b.
PROCEDUREFactoring by FOIL
ax2 + bx + c = (__x + __)(__x + __)
Product must be c.
Product must be a.
PROCEDURE
1. The product of the numbers in the first (F) blanks must be a.
Factoring by FOIL
PROCEDUREFactoring by FOIL
2. The coefficients of the outside (O) products and the inside (I) products must add up to b.
PROCEDURE
3. The product of numbers in the last (L) blanks must be c.
Factoring by FOIL
Section 5.3Exercise #8
Chapter 5Factoring
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 6x2 – 11xy + 3y2 .
3 2 x x – y – 3y
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
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.
RULESFactoring Rules 2 and 3: PERFECT SQUARE TRINOMIALS
X 2 + 2AX + A2 = (X + A)2 (F2)
22 2Note that + ( + )X A X A
RULESFactoring Rules 2 and 3: PERFECT SQUARE TRINOMIALS
X 2 – 2AX + A2 = (X – A)2
22 2Note that – ( – )X A X A (F3)
RULEFactoring Rule 4: THE DIFFERENCE OF TWO SQUARES
X2– A2 = (X + A)(X – A) (F4)
Section 5.4Exercise #11
Chapter 5Factoring
Factor 9x2 – 12xy + 4y2 .
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 9x2 – 12xy + 4y2 .
= 3x – 2y 2
Section 5.4Exercise #13
Chapter 5Factoring
Factor 16x2 – 25y2 .
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 16x2 – 25y2 .
= 4 + 5 4 – 5x y x 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.
RULEFactoring Rule 5: THE SUM OF TWO CUBES.
X 3+ A3 = (X + A)(X 2 – AX + A2)
(F5)
RULEFactoring Rule 6: THE DIFFERENCE OF TWO CUBES. (F6)
X 3– A3 = (X – A)(X 2+ AX + A2)
PROCEDUREGeneral Factoring Strategy
1. Factor out all common factors.
PROCEDUREGeneral Factoring Strategy
2. Look at the number of terms inside the parentheses. If there are:
Four terms: Factor by grouping.
PROCEDUREGeneral Factoring Strategy
Three terms:
If the expression is a perfect square trinomial, factor it. Otherwise, use the ac test to factor.
PROCEDUREGeneral Factoring Strategy
Two terms and squared:
Look at the difference of two squares (X
2–A2) and factor it.
Note: X 2+A2 is not factorable.
PROCEDUREGeneral Factoring Strategy
Two terms and cubed:
Look for the sum of two cubes (X
3+A3) or the difference of two cubes (X
3-A3) and factor it.
PROCEDUREGeneral Factoring Strategy
Make sure your expression is completely factored.
Check by multiplying the factors you obtain.
Chapter 5Factoring
Section 5.5
Section 5.5Exercise #15
Chapter 5Factoring
Factor 8y3 – 125x3 .
3 3 = 2 – 5y x
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 8y3 – 125x3 .
= 2y 3 – 5x 3
= 2y – 5x 4y2 + 10xy + 25x2
Section 5.5Exercise #17
Chapter 5Factoring
Factor 2x3 – 8x2 – 10x.
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 2x3 – 8x2 – 10x.
2 = 2 – 4 – 5x x x
= 2 – 5 + 1x x x
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor 2x3 + 6x2 + x + 3.
2 = + 3 2 + 1x x
2 = 2 + 3 + 1 + 3x x x
3 2 = 2 + 6 + + 3x x x
Section 5.5Exercise #20
Chapter 5Factoring
Factor – 9x4 + 36x2 .
Factor out GCF
(3) Terms(2) Terms (4) Terms
Difference of Squares
Sum/Difference of Cubes
Perfect Square Trinomial
(x2 + bx + c)
(ax2 + bx + c)
Grouping
Factoring Strategy Flow Chart
Factor – 9x4 + 36x2 .
2 2 = – 9 – 4x x
2 = – 9 + 2 – 2x x x
Section 5.6
Quadratic Equations
OBJECTIVES
A Solve quadratic equations by factoring.
DEFINITIONQuadratic Equation in Standard Form
If , and are real numbers (a 0),
a b c
ax2+ bx + c = 0
PROCEDURESolving Quadratics by Factoring
1. Perform necessary operations on both sides so that right side = 0.
PROCEDURESolving Quadratics by Factoring
2. Use general factoring strategy to factor the left side if necessary.
PROCEDURESolving Quadratics by Factoring
3. Use the principle of zero product and make each factor on the left equal 0.
PROCEDURESolving Quadratics by Factoring
4. Solve each of the resulting equations.
PROCEDURESolving Quadratics by Factoring
5. Check results by substituting solutions obtained in step 4 in original equation.
Section 5.6Exercise #24
Chapter 5Factoring
(2x – 3)(x – 4) = 2(x – 1) – 1Solve.
2x2 – 8x – 3x + 12 = 2x – 2 – 1
2x2 – 11x + 12 = 2x – 3
2x2 – 13x + 12 = – 3
2x2 – 13x + 15 = 0
2x – 3 = 0 x – 5 = 0or
(2x – 3)(x – 5) = 0
(2x – 3)(x – 4) = 2(x – 1) – 1Solve.
2x – 3 = 0 x – 5 = 0or
2x = 3 x = 5
x =
32
Section 5.7
Quadratic Equations
OBJECTIVESA Integer problems.
B Area and perimeter problems.
OBJECTIVESC Problems involving the
Pythagorean Theorem.
D Motion problems.
NOTETerminology
Examples: 3,4; – 6,–5
Notation2 consecutive integers
n, n+1
NOTETerminology
Examples: 7, 8, 9; – 4,– 3,– 2
Notation3 consecutive integers
n, n+1, n+2
NOTETerminology
Examples: 8,10; – 6,– 4
Notation2 consecutive even integers
n, n +2
NOTETerminology
Examples: 13,15; – 21,– 19
Notation2 consecutive odd integers
n, n +2
DEFINITIONPythagorean Theorem
a2 + b2 = c2
If the longest side of a right triangle is of length c and the other two sides are of length a and b, then
DEFINITIONPythagorean Theorem
a2 + b2 = c2
Leg a
Hypotenuse c
Leg b
Section 5.7Exercise #26
Chapter 5Factoring
The product of two consecutive odd integers is 13 more than 10 times the larger of the two integers. Find the integers.
Let x + 2 = 2nd odd integer
x (x + 2) = 10(x + 2) + 13
x2 + 2x = 10x + 20 + 13
x2 + 2x = 10x + 33
x2 – 8x = 33
x2 – 8x – 33 = 0
Let x = 1st odd integer
The product of two consecutive odd integers is 13 more than 10 times the larger of the two integers. Find the integers.
x2 – 8x – 33 = 0
(x – 11)(x + 3) = 0
x – 11 = 0 x + 3 = 0 or
x = 11 x = – 3 or
Let x = 1st odd integer
Let x + 2 = 2nd odd integer
The product of two consecutive odd integers is 13 more than 10 times the larger of the two integers. Find the integers.
x = 11 x = – 3 or
x + 2 = 13 x + 2 = – 1
The integers are 11 and 13 or – 3 and – 1.
Let x = 1st odd integer
Let x + 2 = 2nd odd integer
Section 5.7Exercise #29
Chapter 5Factoring
A rectangular 10-inch television screen (measured diagonally) is 2 inches wider than it is high. What are the dimensions of the screen?
x2 + (x + 2)2 = 102
Let x = height
x
Let x + 2 = length
x + 2
10
x2 + (x + 2)2 = 102
x 2 + x2 + 4x + 4 = 100
2x2 + 4x + 4 = 100
2x2 + 4x – 96 = 0
x2 + 2x – 48 = 0
(x + 8)(x – 6) = 0
x + 8 = 0 x – 6 = 0 or x = – 8 x = 6
height –8 x + 2 = 8
The screen is 6 inches high and 8 inches long.
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