# Intermediate Algebra Optional Pre-Final Exam Review

## Presentation on theme: "Intermediate Algebra Optional Pre-Final Exam Review"— Presentation transcript:

Intermediate Algebra Optional Pre-Final Exam Review
1 – Basic Algebra Review 2 – Graphs & Equations of Lines 3 – Solving Systems of Equations 4 – Inequalities 5 – Polynomials & Factoring 6 – Rational Expressions & Functions 7 – Radical Expressions & Functions 8 – Quadratic Functions 9 – Exponents & Logarithms 10 – Conic Sections

1 – Basic Algebra Review

1 – Basic Algebra Review Rules for Order of Operations
To make sure an expression is always evaluated in the same way by different people, the Order of Operations convention was defined Mnemonic: “Please Excuse My Dear Aunt Sally” Parentheses Exponents Multiply/Divide Add/Subtract Always: Evaluate & Eliminate the innermost grouping first

1 – Basic Algebra Review

2 – Graphs & Equations of Lines Plotting Points aka Graphing Points
( , )

2 – Graphs & Equations of Lines Solutions to Equations
Any point on a graphed equation is a Solution

2 – Graphs & Equations of Lines What is Slope & Why is it Important?
Using any 2 points on a straight line will compute to the same slope.

2 – Graphs & Equations of Lines The Slope-Intercept Form of a Line
Remember the CoverUp Method?

Slopes of Parallel Lines m1 = m2
But wait! can you be sure that it’s not the same line ?

Slopes of Perpendicular Lines m1 = -1 / m2

3 – Solving Systems of Equations Using the Substitution Method

3 – Solving Systems of Equations Using the Elimination (Addition) Method

3 – Solving Systems of Equations Solution to 3 Equations
Adding (A) and (C) will eliminate y (A) 2x – y + 3z = 6 (C) 2x + y + z = (D) 4x +4z = first new equation in 2 variables Adding (B) and 5·(C) will also eliminate y (B) 3x – 5y +4z = 7 5·(C) 10x + 5y + 5z = (E) 13x + 9z = second new equation in 2 variables Solve (D) and (E) like a system of two equations (next page) Use Substitution or Addition

Solution to (D) 4x + 4z = 4 Continued (E) 13x + 9z = -3
Well use substitution of x from (D) into (E) to find z (D) 4x +4z = (D1) x = 1 - z move 4z to the other side, divide by 4 Substitute x from (D1) into (E) (E) x + 9z = (1 – z) + 9z = – 13z + 9z = use distribution, then simplify z =  z = 4 Substitute z into (D) or (E) or (D1) to find x (D) 4x + 4(4) = x + 16 = x =  x = -3 Substitute x and z into (A) or (B) or (C) to find y (C) 2(-3) + y + (4) = y + 4 = y = 6 Solution is (-3, 6, 4)

4 – Inequalities Intersections, Unions & Compound Inequalities
Set Diagrams Intersections of Sets Conjunctions of Sentences and Unions of Sets Disjunctions of Sentences or Interval Notation Domains

4 – Inequalities Expressing Domains With Interval Notation

4 – Inequalities Using the Absolute Value Principle
|x + 1| = 2 x + 1 = 2 or x + 1 = -2 |2y – 6| = 0 2y – 6 = 0 |5x – 3| = -2 no solution

4 – Inequalities When an equation has 2 absolute values?

5 – Polynomials & Factoring Subtracting Polynomials
To subtract polynomials, add the opposite of the second polynomial. (7x3 + 2x + 4) – (5x3 – 4) add the opposite! (7x + 2x + 4) + (-5x3 + 4) Use either horizontal or vertical addition. Sometimes the problem is posed as subtraction: x2 + 5x make it addition x2 + 5x (x2 + 2x) _ of the opposite x2 – 2x__ x +6

5 – Polynomials & Factoring Multiplying Two Polynomials
To multiply a polynomial by a polynomial, we use the distributive property repeatedly. Horizontal Method: (2a + b)(3a – 2b) = 2a(3a – 2b) + b(3a – 2b) = 6a2 – 4ab + 3ab – 2b2 = 6a2 –ab – 2b2 Vertical Method: 3x2 + 2x – 5 4x + 2 6x2 + 4x – 10 12x3 + 8x2 – 20x____ 12x3 + 14x2 – 16x – 10

5 – Polynomials & Factoring FOIL: Used to Multiply Two Binomials

5 – Polynomials & Factoring Find the Greatest Common Factor
7(? – ?) = 7(a – 3) 19x3 + 3x = x(? + ?) = x(19x2 + 3) 18y3 – 12y2 + 6y = 6y(? – ? + ?) = 6y(3y2 – 2y + 1)

5 – Polynomials & Factoring Factor by Grouping
8t3 + 2t2 – 12t – 3 2t2(4t + 1) – 3(4t + 1) (4t + 1)(2t2 – 3) 4x3 – 6x2 – 6x + 9 2x2(2x – 3) – 3(2x – 3) (2x – 3)(2x2 – 3) y4 – 2y3 – 12y – 3 y3(y – 2) – 3(4y – 1) Oops – not factorable via grouping

5 – Polynomials & Factoring Using a Factor Table - Trial & Error
Let’s use x2 + 13x as an example Factors must both be sums: (x + ?)(x + ?) Pairs=c= Sum=b=13 1, 2, 3, 4, ok quit! x2 + 13x + 36 = (x + 4)(x + 9)

The ac Grouping Method: ax2 + bx + c Split bx into 2 Terms: Use a Table based on a·c
Let’s use 3x2 – 10x – 8 as an example ac = 3(-8) = -24 One factor is positive, the other negative and larger. Pairs=ac= Sum=b=-10 1, 2, quit! 3, 4, 3x2 – 10x – 8 = 3x2 + 2x – 12x – 8 = split the middle x(3x + 2) – 4(3x + 2) = do grouping = (3x + 2)(x – 4)

5 – Polynomials & Factoring Factoring Perfect Square Trinomials?
x2 + 8x + 16 = (x + 4)2 (x) (4) (x)(4) = 8x yes, it matches t2 – 12t + 4 = not a PST (t) (-2) (t)(-2) = -4t no, it’s not -12t 25 + y2 + 10y = (y + 5)2 y2 + 10y descending order (y) (5) (y)(5) = 10y yes, it matches 3x2 – 15x + 27 = not a PST 3(x2 – 5x + 9) remove common factor (x) (-3) (x)(-3) = -6x no, it’s not -5x PST Tests: 1. Descending Order 2. Common Factors 3. 1st and 3rd Terms (A)2 and (B)2 4. Middle Term 2AB or -2AB

5 – Polynomials & Factoring Difference of Squares Binomials
Remember that the middle term disappears? (A + B)(A – B) = A2 - B2 It’s easy factoring when you find binomials of this pattern A2 – B2 = (A + B)(A – B) Examples: x2 – 9 = (x)2 – (3)2 = (x + 3)(x – 3) 4t2 – 49 = (2t)2 – (7)2 = (2t + 7)(2t – 7) a2 – 25b2 = two variables squared (a)2 – (5b)2 = (a + 5b)(a – 5b) 18 – 2y4 = constant 1st, variable square 2nd 2 [ (3)2 – (y2)2 ] = 2(3 + y2)(3 – y2)

5 - Factoring the Difference between 2 Cubes X3 – Y3 = (X – Y)(X2 + XY + Y2)
F3 – L3 factors easily to (F – L)(F2 + FL +L2) Examine 27a3 – 64b3 (3a)3 – (4b)3 (3a – 4b)(9a2 + 12ab + 16b2) Remember to remove common factors and to factor completely p3 – x6 – 128 = 2[x6 – 64] (p)3 – (2) [(x2)3 – 43] (p – 2)(p2 + 2p + 4) (x2 – 4)(x4 + 4x2 + 16) (x + 2)(x – 2)(x4 + 4x2 + 16)

5 - Factoring the Sum of 2 Cubes X3 + Y3 = (X + Y)(X2 – XY + Y2)
F3 + L3 factors easily to (F + L)(F2 – FL +L2) Examine 27a3 + 64b3 (3a)3 + (4b)3 (3a + 4b)(9a2 – 12ab + 16b2) Remember to remove common factors and to factor completely p x = 2[x6 + 64] (p)3 + (2) [(x2)3 + 43] (p + 2)(p2 – 2p + 4) (x2 + 4)(x4 – 4x2 + 16)

Principle of Zero Factors
5 - Definition Principle of Zero Factors

5 - Solving a Quadratic Equation by Factoring

6 – Rational Expressions & Functions Multiplying Fractions
(Use parentheses for clarity) Factor expressions, then cancel like factors

6 – Rational Expressions & Functions Dividing Fractions
Change Divide to Multiply by Reciprocal, follow multiply procedure

6 - Finding the LCD (must be done before adding or subtracting 2 or more RE’s)
1. Factor each denominator completely into primes. 2. List all factors of each denominator. (use powers when duplicate factors exist) 3. The LCD is the product of each factor to its highest power. 28z3 = (22) (7)(z3) 21z = (3)(7)(z) LCD=(22)(3)(7)(z3) a2 – = (a + 5)(a – 5) a + 7a + 10 = (a + 5) (a + 2) LCD = (a + 5)(a – 5)(a + 2)

6 - Adding or subtracting rational expressions with unlike denominators
1. Find the LCD. 2. Express each rational expression with a denominator that is the LCD. 3. Add (or subtract) the resulting rational expressions. 4. Simplify the result if possible.

6 - Simplifying Complex Rational Ex’s Method 1: Multiplying by 1

6 - Simplifying Complex Rational Ex’s Method 2: Multiplying by Reciprocal
Making the top and bottom into single expressions, then multiplying by reciprocal.

6 – Rational Expressions & Functions Rational Equations: False Solutions
Solve a Rational Equation by Multiplying BOTH SIDES by the LCD Warning: Clearing an equation may add a False Solution A False Solution is one that causes a divide by zero situation in the original equation Before even starting to solve a rational equation, we need to identify values to be excluded What values need to be excluded for these? t ≠ a ≠ ± x ≠ 0

6 – Rational Expressions & Functions Clearing & Solving a Rational Equation
What gets excluded? x ≠ 0 What’s the LCD? 15x What’s the solution?

6 – Rational Expressions & Functions Dividing a Polynomial by a Polynomial
Use the long-division process

7 – Radical Expressions & Functions Examples to Simplify

7 – Radical Expressions & Functions Practice
Express using rational exponents: Simplify using rational exponents:

Rational Expressions Where the Numerator is greater than 1
Using Exponent Arithmetic, it’s a little easier

7 – Radical Expressions & Functions A Radical Expression is Simplified When:
Each factor in the radicand is to a power less than the index of the radical The radicand contains no fractions or negative numbers No radicals appear in the denominator of a fraction

7 – Radical Expressions & Functions Definitions
A Radical Equation must have at least one radicand containing a variable The Power Rule: If we raise two equal quantities to the same power, the results are also two equal quantities If x = y then xn = yn Warning: These are NOT equivalent Equations!

To eliminate the radical, raise both sides to the index of the radical

Make sure radicals are on opposite sides Sometimes you need to repeat the process

7 – Radical Expressions & Functions i, The Basis of the Complex Number System

7 – Radical Expressions & Functions Powers of i

8 – Quadratic Functions The Square Root Principle
Solve by factoring x2 – 16 = (x+4)(x-4)=0 x=4,-4 Then by the square root property x2 – 16 = x2 = x=4,-4

8 – Quadratic Functions Using Completing the Square to Solve an Equation

The Quadratic Formula is used to find solutions to any quadratic equation The formula was derived using completing the square and the square root property.

The variable u is often used to replace squares of variables or expressions

8 – Quadratic Functions (parabolas) Graph f(x) = (x – 3)2

Maximums & Minimums

9 – Exponents & Logarithms Graphing an Exponential Function

9 – Exponents & Logarithms Translating Right and Left

9 – Exponents & Logarithms General Definition of Logarithms

10 - Introduction to Conic Sections Parabola Circle Ellipse Hyperbola

10 – Conic Sections Remember Parabolas?
Two styles: Functions & Relations Find the Vertex: x = -b/(2a), (or y = -b/(2a)) solve for y or x

10 – Conic Sections A Circle has a Center and a Radius