1. 5.1
A quadratic function f is a function of the form f(x) = ax2 + bx + c where a, b and c are real numbers and a not equal to zero. The graph of the quadratic function is called a parabola. It is a "U" shaped curve that may open up or down depending on the sign of coefficient a.
Jeff White

2. Graphing A Quadratic Function
Vertex is (2,-2)
Then draw the axis of symmetry which is x=2
Then plot two points on one side of the axis of symmetry.
Use symmetry to plot two more points.
Jeff White

3. Graphing A Quadratic Function In Vertex Form
First plot the vertex (H,K) = (-3,4)
Then draw the axis of symmetry X=-3 and plot two points on one side of it.
Use symmetry to complete the graph.
Jeff White

4. Graphing A Quadratic Function In Intercept Form
X-intercept occur at (-2,0) and (4,0)
Axis of symmetry is 1
X-coordinate of the vertex is x=1.
The y-coordinate of the vertex is
Jeff White

5. 5.4 Complex Numbers Imaginary unit is called i i= √ (-1)
r is a positive real number
√ (–r)= i √ (r)
Complex number written in standard form is a+bi
a & b real numbers
a real part of complex number
b imaginary part of complex number
b ≠0
a+bi is imaginary
a=0, b≠0
a=bi is pure imaginary number
z=a+bi is complex number

6. Sample Problems
2. Write (8+5i)-(1+2i) as a complex number in standard form.
4. Multiply i(3+i).
6. Find the absolute value of 3-4i
1. Solve
3. Write the (2+3i)+(7+i) as a complex number in standard form.
5. Divide

7. Helpful Hints a=3, b=-4 1. Take the square root of x squared and -4.
2. Distribute the minus sign to 1 and 2i. Combine like terms.
3. Distribute the plus sign to 7 and i. Combine like terms.
4. Distribute the i to 3 and i.
5. Divide 8 by 1 and 8 by i.
6. Consult the formula on the first page:
a=3, b=-4

8. Answers 1. 2.

9. 4. 3.

10. 5. 6.

11. Chapter 5.5: Completing the Square
Goal 1: Solving Quadratic Equations by Completing the Square Completing the square is a process that allows you to write an expression of the form x2 + bx as the square of a binomial. To complete the square for x2 + bx, you need to add (b/2) 2. The following is a rule for completing the square: X2 + bx +(b/2)2 = (x+[b/2])2
Example 1: Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
x2 + 18x + c Write the equation out
b= Use the formula to find b
c = (b/2)2 = (18/2)2 = 92 = 81 Find the value of c that makes the expression a perfect square trinomial
x2 + 18x Substitute the C value in the expression.
(x+9)2 Factor to get your answer
Matt

12. Chapter 5.5: Completing the Square
Example 2: Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square.
X2 + 2x = 9 Write out original equation
X2 + 2x + 1 = Add (2/2)2 = 12 = 1 to each side
(x+1)2 = Write the left side as a binomial squared
X + 1 = √10 Take the square roots of each side
X = -1 + √10 Solve for x
Example 3: Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square.
6x2 +84x +300 = 0 Write the original equation
X2 +14x +50 = 0 Divide both sides by the coefficient of x2
X2 + 14x = -50 Write the left side in the form of x2 + bx
X2 + 14x + 49 = -1 Add (14/2)2 = 72 = 49 to each side
(X + 7)2 = -1 Write left side as a binomial squared
X + 7 = √-1 Take the square roots of each side
X = -7 ± √-1 Solve for x
X = -7 ± i Write in terms of the imaginary unit i

13. Chapter 5.5: Completing the Square
Goal 2: Writing Quadratic Functions in Vertex Form: Given a quadratic function in standard form, y = ax2 + bx + c, you can use completing the square to write the function in vertex form, y = a(x – h)2 + k.
Example 4: Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex.
Y = x2 – 6x Write out the original function
Y + 9 = (x2 – 6x + 9) Complete the square of x2 – 6;
add (-6/2)2 = -32 = 9
Y + 9 = (x – 3) Write x2 – 6x + 9 as a binomial squared
Y = (x – 3) Solve for y
Vertex = (3,2)

14. Chapter 5.5: Completing the Square
Practice Problems for Completing the Square: Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. X2 – 44x + c Answer: C = 484; (x – 22)2 Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is 1: Solve the equation by completing the square. X2 + 20x = 0 Answer: i, -10 – 2i Practice Problems for Solving a Quadratic Equation if the Coefficient of x2 Is Not 1: Solve the Equation by Completing the Square. 2x2 – 12x = -14 Answer: 3 ± √2 Practice Problems for Writing a Quadratic Function in Vertex Form: Write the quadratic function in vertex form and identify the vertex. Y = x2 – 3x – 2 Answer: y = (x – [3/2])2 – (17/4) Vertex: ([3/2], [-17/4])

15. A presentation by: Elise Couillard; Block 5 June 9, 2010
Algebra II Section 5.6
A presentation by:
Elise Couillard; Block 5
June 9, 2010

16. Solving Equations With The Quadratic Formula
By completing the square once for the general equation , you can develop a formula that gives the solutions of any quadratic equation. The formula for the solutions is called the quadratic formula.

17. The Quadratic Formula The Quadratic Formula:
*Let a, b, and c be real numbers such that a does not equal 0. The solutions of the quadratic
equation are:

18. Solving a Quadratic Equation With 2 Real Solutions
Solve
The solutions are x=1.35 and x=-1.85

19. Number and Type of Solutions of a Quadratic Equation
Consider the quadratic equation
*If >0, then the equation has two real solutions.
*If =0, then the equation has one real solution.
*If <0, then the equation has two imaginary solutions.

20. Sources: Algebra II Textbook
The End!

21. Lesson 5.7 Jessica Semmelrock
Quadratic Inequality
in 2 Variables
Quadratic Inequality
in 1 Variable
y < ax2 + bx +c
y < ax2 + bx +c
ax2 + bx +c < 0
ax2 + bx +c < 0
y < ax2 + bx +c
y > ax2 + bx +c
ax2 + bx +c > 0
ax2 + bx +c > 0
Example 1
1. Graph y < x2 + 8x + 16
2. Test the point (0,0)
y < x2 + 8x + 16
0 < (0)+ 16
0 < 16
3. Shade outside region because 0 < 16
Lesson 5.7 Jessica Semmelrock

22. Example 2: Graph the system of quadratic inequalities y > x2
Only difference to this problem is when the shading overlaps that is your answer.
Lesson 5.7 Jessica Semmelrock

23. Example 3: Solve x2 + x -2 < 0 by graphing Step 1: x2 + x -2 = 0
Find graph intercepts by replacing 0 for x.
Step 2: x = – 4(1)(-2)
2(1)
Use quadratic formula to solve for x.
Answer: x ≤ –1.37 or x ≥ .37
Lesson 5.7 Jessica Semmelrock

24. Example 4: Answer x = 3 or x = -6
Solve x > 0 algebraically
x > 0
x = 0
(x-3) (x+6) = 0
Test an x-value in each interval to see if it satisfies the inequality.
Test these points with the arrows.
Answer x = 3 or x = -6
Lesson 5.7 Jessica Semmelrock

25. Travis Deskus
6.1 / 7.2

26. Using Properties of Exponents-
Using Properties of Exponents

27. Evaluating Numerical Expressions
1. Product of like bases: Example: x5 x3 = x5+3 = x8
To multiply powers with the same base, add the exponents and keep the common base.
=32
2.Evaluating numerical expresions a.

28. Simplifying Algebraic Expressions
Scientific Notation

29. 7.2 Properties of Rational Exponents
Example 1.  
If m= for some integer n greater than 1, the third and sixth properties can be written using radical notation as following:
product property
Quotient property a. b.
Using Properties of Radicals

30. Writing Radicals in Simplest Form
a. steps- factor out perfect cube, product property, simplify
Adding and Subtracting Roots and Radicles
Book Example
The properties of rational exponent and radicals can also be applied to expressions involving variables. Because a variable can be positive, negative or zero, sometimes absolute value is needed when simplifying a variable expression.
=x when n is odd = when n is even
Simplifying expressions involving varribles

31. Section 6.2

32. IDENTIFYING POLYNOMIAL FUNCTIONS
A polynomial is a function if it’s in standard form and the exponent is a whole number
ex: f(x)= 3
If the polynomial has an exponent that is not a whole number it’s not a function
ex: f(x)= 3x1/2 – 2x2 +5

33. Using Synthetic Substitution
Write the polynomial in standard form
Insert terms with coefficients of 0 for missing terms
Then write the coefficients of f(x) in a row
Bring down the leading coefficients and multiply them by 1
Write results in the next column and bring down your results
Continue until you reach the end of the row
EX:

34. Graphing Polynomials Functions
Begin by making a table of values, including positive, negative, and zero values for x
Plot the points and connect them with a smooth curve. Then check the behavior
The Degree is odd and the leading coefficients is positive, so
f(x)→ + -
As x → -
f(x) → +
As x → + X -3 -2 -1 1 2 3
f(x)
-218
-18 6 4 42
262

35. Examples of Graphs

36. Work Graph the polynomial function 1.) f(x)= x4+3 2.) g(x)= x3-5
3.) h(x)= 2+ x2-x4
Synthetic substitution
1.) f(x)=x3+ 5x2+4x+6, x=2
2.) f(x)= x3-x5 +3, x=-1
3.) f(x)= 5x3-4x2-2, x=0
Functions yes or no
2.) f(x)=5x3/4-5x2+3

37. Adding, Subracting, and Multiplying Polynomials
6.3
By: Robert Johnson

38. How to Solve
To add or subtract polynomials, add or subtract the coefficients of LIKE terms. You can do this by using a vertical or horizontal format
To Multiply two polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial. Then combine LIKE terms

39. Adding Polynomials Add 2x3-5x2+3x-9 and x3+6x2+11 in vertical format
Add 3x3+2x2-x-7 and x3-10x2+8 in horizontal format

40. Subtracting Polynomials
Subtract 3x3+2x2-x+7 from 8x3-x2-5x+2 in vertical format
Subract 8x3-3x2-2x+9 from2x3+6x2-x+1 in horizontal format

41. Multiplying Polynomials
Multiply -2y2+3y-6 and y-2 in vertical format
Multiply -x2+2x+4 and x-3 in horizontal format

42. Special Product Patterns
Sum and difference
(a+b)(a-b)= a2 - b2
Square of a Binomial
(a+b)2 = a2 + 2ab + b2
(a-b)2 = a2 - 2ab + b2
Cube of a Binomial
(a+b)3 = a3 + 3a2b + 3ab2 + b3
(a-b)3 = a3 - 3a2b + 3ab2 - b3

43. Using Special Product Patterns
(x + 2)(3x2 - x – 5)
(a – 5)(a + 2)(a + 6)
(xy - 4)3

44. 6.4 Factoring and solving polynomials
Anjy Grasso

45. Step One Step Two x² + 8x + 12 = 0 x² + 8x = −12 x² + 8x + 12 = 0
Now that you have the equation you're solving, find the first factor
Put the equation into Standard form
In other words, make it equal zero.
x² + 8x = 0
( ) ( ) = 0
x² + 8x = −12
the only factors of x2 are x * x, you now have the first factors.
x² + 8x = 0
x² + 8x = 0
(x ) (x ) = 0
Anjy Grasso

46. Now we find the factors of 12.
Step Three
Now that we have the first factors, the x2 goes away, and we're left with this:
8x = 0
(x ) (x ) = 0
8x = 0
(x ) (x ) = 0
(x 6) (x 2) = 0
Now we find the factors of 12.
Since the entire equation was positive,
Both of these should be positive too.
1 and 12 wont work,
neither will 4 and 3,
so lets use 2 and 6
(x + 6) (x + 2) = 0
Anjy Grasso

47. So how exactly do we know this is right?
Checking Your Work
So how exactly do we know this is right?
Lets use FOIL (first, outer, inner, last) multiplication to test it out.
(x + 6) (x + 2) = 0
F - First
x * x = x2
O - Outer
x * 2 = + 2x
x2 + 2x + 6x + 12
I - Inner
6 * x = + 6x
Now Simplify…
x2 + 8x + 12
L - Last
6 * 2 = + 12
There's Your original equation!
Anjy Grasso

48. 6.5 Remainder and Factor Theorems
Polynomial Long Division – when dividing a polynomial f(x) by a divisor d(x), you get a quotient polynomial
-q(x) and a remainder polynomial r(x) f(x) = q(x) + r(x)
d(x) d(x)
Remainder Theorem – If polynomial f(x) is divided by x-k, then the remainder is r = f(x)
Synthetic Division – only use the Coefficients of the polynomial and the x – k must be in the form of a divisor.
Factor Theorem – A polynomial f(x) has a factor x-k if and only if f(k) = 0
Rachael SKinner

49. Brief Refresher f(x) = 3x² + 4x -3 by x² - 3x + 5 - -
Long Division
f(x) = 3x² + 4x -3 by x² - 3x + 5
* Don't forget to add in exponents if needed exponents must go in numerical order
3x² + 4x -3
x² - 3x + 5
3x⁴ - 5x³ 0x² + 4x – 6 -
3x⁴- 9x³+ 15x²
4x² - 15x² + 4x -
4x² - 12x² - 20x
* Remember to subtract – which means the signs will change
-3x² - 16x-6
-3x² + 9x -15
25x+9
25x+9
Remainder
3x² + 4x -3
x² - 3x + 5
Rachael SKinner

50. Brief Review f(x) = 2x³ +x² - 8x +5 by x + 3 -3 Synthetic Division
Factoring Completely
f(x) = 3x³ - 4x² - 28x – 16
x +2 is a factor
x + 3 = 0
x – 3 = - 3
x = - 3
x = 2 2
*Use only coefficients
no remainder -3
f(x) (2 + x) (3x² - 10x -8)
Factor
Remainder
f(x) (x+2) (3x+2) (x – 4)
Solve
finding the 0's of f(x)
(x+2) = -2
(3x+2) = -2/3
(x – 4) = 4
2x³ +x² - 8x
x+3
Rachael SKinner

51. Practice Problem 1.) Divide 2x⁴ + 3x³ + 5x – 1 by x² – 2x +2
Using Long Division
2.) Divide x³ + 2x² - 6x – 9 by x-2
Using Synthetic Division
3.) Factor Completely given that x-4 is a factor
f(x) = x³ + 6x² + 5x +12
Rachael SKinner

52. Answers x² -2x +2 2.) x² + 4x + 2 + -5 x- 2 1.) 2x² + 7x +10 + 11x-21
3.) f(x) = (x-3) (x-4) (x+1)
Rachael SKinner

53. How to find the rational zeros of a polynomial function
Lesson 6.6
Nate

54. Rational Zero Theorem
If f(x)=anxn+…..+a1x+a0 has integer coefficients, then every rational zero of f has the form: P/Q =
factor of constant term a0 / factor of leading coefficient an
Nate

55. Steps to Solve
List all possible rational zeros, both positive and negative.
Change the number that is outside until one of these numbers makes your answer zero.
When your answer is zero, insert the numbers you got in order to find the value of x.
Nate

56. Example
Solve: f(x)=3x3-4x2-17x+6
Nate

57. Answer 3 -4 -17 6 Synthetic Division -2 -6 20 -6
x= 1,2,3,6 (all #’s are +or -) /
1,3 (all #’s are + or -)
f(x)=(x+2)(3x2-10x+3) Factor the trinomial
f(x)=(x+2)(3x-1)(x-3) and use the factor x=-2,3,1/3 theorem
Nate

58. 7.1 nth roots n is an integer greater than 1 and a is a real number:
n is odd, a has one nth root: =a
n is even and a>0 a has 2 nth roots: = a
n is even and a=0 a has 1 nth root: =0 =0
n is even and a<0 a has 0 nth roots
Loren

59. 7.1 nth roots a is the nth root of a and m is a positive integer:
a =(a ) m =( ) m
a = = m = m , a 0
Loren

60. Practice Problems Rewrite using rational exponent notation.
Fine the indicated real nth root(s) of a
n=4, a=0
Evaluate the expression
Loren

61. Power Functions and Function Operations
Section 7.3
Power Functions and Function Operations
Andy

62. Examples 1. Combine the coefficients 1. Combine the coefficients
= 4x1/2 + (–9x1/2)
= 4x1/2 – (–9x1/2)
1. Combine the coefficients
1. Combine the coefficients
= [4 + (–9)]x1/2
= [4 – (–9)]x1/2
2. Simplify
2. Simplify
= 13x1/2
= –5x1/2
Andy

63. Practice Problems
f (x) + g(x)
2. f (x) – g(x)
Andy

64. Examples b. f (x) g(x) a. f (x) g(x) Plug in equation Plug in equation
= (6x)(x3/4) = 6x
x3/4
Combine powers by adding
Combine powers by subtraction
= 6x(1 + 3/4)
Simplify =
6x(1 – 3/4)
Simplify
= 6x7/4 =
6x1/4
Andy

65. Practice Problems
f (x) g(x)
f (x)
g(x)
Andy

66. Examples a. f(g(x)) b. g(f(x)) Plug in equation Plug in equation=
f(5x – 2)
Plug g(x) into the equation
Plug g(x) into the equation
= 5(4x–1) – 2
= 4(5x – 2)–1
Distribute the 5
Plug in the coefficient and power of f(x)
= –20x–1 – 2
Simplify 20 x
– 2 =
Andy

67. Examples c. f(f(x)) Plug in equation = f(4x–1)
Plug f(x) into the equation
= 4(4x–1)–1
Distribute the exponent
= 4(4–1 x)
Distribute the 4
= 40x
= x OR
Andy

68. Practice Problems a.
f(g(x)) b.
g(f(x)) c.
f(f(x))
Andy

69. 7.4 Inverse Functions Concepts
Inverse relations – maps the output values back to their original input values
Domain – of inverse relation is the range of the original relation.
Range – of inverse relation is the domain of the original relation.
Inverse of linear functions: reflect original relation over y=x to obtain the inverse relation.
Which means switch the roles of x & y and solve for y (if possible).
Ileishka Ortiz

70. 7.4 Original relation Inverse relation DOMAIN DOMAIN RANGE RANGE
Ileishka Ortiz

71. 1.) Find an equation for the inverse of the relation y = 2 x – 4
SOLUTION
y = 2 x – 4
Write original relation.
x = 2 y – 4
Switch x and y
Add 4 to each side.
x + 4 = 2 y
x + 2 = y 1 2
Divide each side by 2.
Ileishka Ortiz

72. 7.4 Functions f and g are inverses of each other provided:
f (g (x)) = x and g ( f (x)) = x
The function g is denoted by f – 1, read as “f inverse.”
Given any function, you can always find its inverse relation
by switching x and y.
For a linear function f (x ) = mx + b where m  0, the
inverse is itself a linear function.
Ileishka Ortiz

73. 7.4 g (f (x)) = g (2x – 4) = (2x – 4) + 2 = x – 2 + 2 = x 1 2
1.) Verify that f (x) = 2 x – 4 and g (x) = 1/2 x + 2 are inverses.
SOLUTION
Show that f (g (x)) = x and g (f (x)) = x.
g (f (x)) = g (2x – 4)
= (2x – 4) + 2
= x – 2 + 2
= x 1 2
f (g (x)) = f 1/2 x + 2 ( )
= 2 1/2 x – 4
= x + 4 – 4
= x
Ileishka Ortiz

74. 7.4
PRACTICE
Ileishka Ortiz

75. 7.5 + 7.6 With Da Gawd By: Jon Reyyashi
“If Da Gawd Can Do It Perfect, You Can Do It Almost Perfect Too” - Mrs. Delaney

76. Its Easy
To graph y=a√x-h + k or y=a^3√x-h + k
y=a√x 0r y=a^3√x

77. H + K
H means shift the graph left our right and K means move the point either up or down.

78. 7.5 LET US START WITH 7.5 SHALL WE 1. √X+1-3
SO LETS SAY IT IS √X-(-1)+(-3). H=-1 AND K= -3’
TO OBTAIN THE GRAPH OF Y=√X+1-3, SHIFT THE GRAPH OF Y=√X LEFT 1 UNIT AND DOWN 3 UNITS

79. GRAPHING IT EXAMPLE - Y=-3√X-2 + 1 DRAW THE GRAPH OF Y=-3√X
DASHED LINE IN GRAPH
IT BEGINS AT THE ORIGIN AND PASSES THROUGH (1, -3)
YOU THEN SHIFT THE GRAPH 2 UNITS UP AND UP 1 UNIT.
THE GRAPH STARTS (2,1) AND PASSES THROUGH POINT (3, -2)

80. Domain and Range State the domain and range of the functions
From the graph of y=-3√x the domain is x≥2 and the range is y≤1

81. Another Example ANOTHER EXAMPLE 3^3√X+2 - 1
1. Sketch the graph of y=3^3√x+2 - 1
This means h=-2 and k=-1
Sketch the graph of y=3^3√x
It passes the origin and points (-1, -3) and (1, 3)
You then shift the graph to the left two and down one
You repeat these steps for each point.
Example - (0,0) becomes (-2, -1) because you basically move the point over -2 and down one so it becomes (-2, -1)

82. Domain and Range
Domain and range are both all real numbers

83. 7.6 Powers Property of Equality: If a=B then A^n = B^n
This means that you can raise each side of an equation to the same power
An extraneous solution is a trial solution that does not satisfy the original equation

84. Example √x + 5 = 9 (√x)^2 = 4^2 x = 16
Isolate the radical by subtracting 5 by each side
(√x)^2 = 4^2
Simplify
x = 16

85. Radical Exponents Example 3x^3/4 = 192
Isolate the power by dividing each side by 3
(x^3/4) = 64
Raise each side by 4/3 power, and cancel the original 4/3 power by performing its reciprocal
x = (64^1/3)4
Apply Properties of roots
x = 4^ 4 = 256
Simplify (The solution is 256) Check answer by substituting.

86. With Only One Radical 3√8x+3 - 5 = -2
Isolate the radical by adding 5 to each side
3√8x+3 = 3^3
Cube each side
8x +3 = 27
Subtract each side of by 3
8x=24
Simplify
x=3

87. Two Radicals 3√2x+4 = 2^3√3-x Cube each side (3√2x+4)^3 = (2^3√3-x)^3
Simplify
2x + 4 = 8(3-x)
Distribute
2x + 4 = x ( Add 8 x to each side
10x + 4 = 24
x=2

88. Follow These Rules and
You will be GAWDLY like Me and get a 100 on the Final. Thank You :)

89. Y = p (x) / q (x) = amxm+am-1xm-1+…+ax+a0 bnxn+bn-1xn-1+…+bx+b0
Section 9.3
Y = p (x) / q (x) =
amxm+am-1xm-1+…+ax+a0
bnxn+bn-1xn-1+…+bx+b0
Sydney

90. Section Concepts
P (x) and q (x) are polynomials with no common factors other than 1.
The graph of the function p (x) /q (x) has the following characteristics
-the x intercepts are the real zeros of p (x) *you set the polynomial p (x) equal to zero and solve*
-there is a vertical asymptote at each real zero of q (x) *you set the polynomial q (x) equal to zero and solve*
-at most there is one horizontal asymptote
Sydney

91. Section 9.3 – Concepts cont.
If m < n, the line y=0 is the horizontal asymptote
If m = n, the line y = Am / Bn is the horizontal asymptote
If m > n, the graph has no horizontal asymptote.
Sydney

92. Section 9.3 – Example 1 (m<n)
Answer:
P (x) has no real zeros, so no x-intercepts
The denominator has no real zeros, so no vertical asymptote
M<n, so the horizontal asymptote is y = 0
Graph:
y = 4
x2 + 1
Sydney

93. Section 9.3 – Example 2 (m=n)
Graph:
y = 3x2
x2-4
Answer:
The numerator’s only zero is zero. X-intercept is (0,0)
Vertical asymptote’s at 2 and -2
M=N so horizontal asymptote is at 3
Sydney

94. Section 9.3 – Example 3 (M>N)
Graph:
y = x2-2x-3
x+4
Answer:
X-intercepts are 3 and -1
Vertical asymptote is -4
M>n, so no horizontal asymptote
Sydney

95. Multiplying and Dividing Rational Expressions
9.4
Multiplying and Dividing Rational Expressions
Brett Robinson

96. Let a, b, and c be nonzero real numbers or variable expressions
Let a, b, and c be nonzero real numbers or variable expressions. Then the following property applies:
Divide out common Factor c
Brett Robinson

97. Multiplying Factor Numerator and Denominator
Divide out common factors before multiplying if possible
Simplified Form
Brett Robinson

98. Dividing Multiply by the reciprocal Factor Divide out common Factors
Simplified Form
Brett Robinson

99. Addition, Subtraction, and Complex Fractions
Section 9.5
Addition, Subtraction, and Complex Fractions
Adding and Subtracting with Like Denominators:
Solve the following problems:
Step I: Add numerators and simplify the expressions
Step I: Subtract numerators
Jorge Verde

100. Adding with Unlike Denominators
Solution:
Step I: First find the LCD of and
The LCD is
Step 2: Use this to rewrite each expression.
Jorge Verde

101. Subtracting with Unlike Denominators
Solution
Jorge Verde

102. Simplifying a Complex Fraction
Solution
Step I: Add fractions in denominator.
Step 2: Multiply by reciprocal.
Step 3: Divide out common factor.
Step 4: Write in simplified form.
Jorge Verde

103. Solving Rational Equations
Section 9.6
Solving Rational Equations
Morgan Hillhouse

104. Different ways to Solve Rational Expressions
Cross Multiplying
Multiple Rational Expressions
Set up your equation.
Multiply together the top expression on the left with the bottom on the right and the top expression on the right with the bottom of the left.
Set the result equal to zero.
Solve for each part.
Set up your equation.
Multiply each side if the equation by the LCD of both terms.
Simplify each side.
Set the simplified form to zero and solve for each result.
Check for extraneous solutions.
Morgan Hillhouse

105. Cross Multiplying X 3 4X=3X-12 ― = ― X-4 4 -3X 4X=-12 4 X=-3
― = ― X
4X=3X-12
-3X
4X=-12 4
X=-3
Morgan Hillhouse

106. Continued Problem #2 8X+48=7X X+6 7 -48 -7X ―― = ― X 8 X=-48
―― = ― X
-48 -7X
X=-48
Morgan Hillhouse

107. Multiple Rational Expressions
2X X X+4
―― ―― = ――――
X X X₂-4X-4
2X X X+4
―― ―― = ――――
X X (3X+2)(X-2)
(3X+2)
(X-2)
CONTINUED ON NEXT SLIDE
(3X+2)
(X-2)
Morgan Hillhouse

108. Continued 6X₂+4X-4X₂+9X-2=17+4 2X(3X+2)-(4X-1)(X-2)=17X+4 -17-4
X=3 or -1
Morgan Hillhouse