Review Topics (Chapter 0 & 1)

Review Topics (Chapter 0 & 1) • Exponents & Radical Expressions • Factoring • Quadratic Equations • Rational Expressions • Rational Equations & Clearing Fractions • Radical Equations • Solving Inequalities • Linear Graphing and Functions • Function Evaluation • Slope and Average Rate of Change • Difference Quotient

Review of Exponents 82 =8 • 8 = 64 24 = 2 • 2 • 2 • 2 = 16 x2 = x • x x4 = x • x • x • x Base = x Base = x Exponent = 2 Exponent = 4 Exponents of 1 Zero Exponents Anything to the 1 power is itself Anything to the zero power = 1 51 = 5 x1 = x (xy)1 = xy 50 = 1 x0 = 1 (xy)0 = 1 Negative Exponents 5-2 = 1/(52) = 1/25 x-2 = 1/(x2) xy-3 = x/(y3) (xy)-3 = 1/(xy)3 = 1/(x3y3) a-n = 1/an 1/a-n = an a-n/a-m = am/an

Raising Quotients to Powers a n b an bn a -n b a-n b-n bn an b n a = = = = Examples: 3 2 32 9 4 42 16 = = 2x 3 (2x)3 8x3 y y3 y3 = = 2x -3 (2x)-3 1 y3 y3 y y-3 y-3(2x)3 (2x)3 8x3 = = = =

Product Rule am• an = a(m+n) x3• x5 = xxx • xxxxx = x8 x-3 • x5 = xxxxx = x2 = x2 xxx 1 x4 y3 x-3 y6 = xxxx•yyy•yyyyyy = xy9 xxx 3x2 y4 x-5• 7x = 3xxyyyy • 7x = 21x-2 y4 = 21y4 xxxxx x2

Quotient Rule am = a(m-n) an 43= 4 • 4 • 4 = 41 = 4 43 = 64 = 8 = 4 42 4 • 4 42 16 2 x5 = xxxxx = x3x5 = x(5-2) = x3 x2 xx x2 15x2y3 = 15 xx yyy = 3y215x2y3 = 3 • x -2 • y2 = 3y2 5x4y 5 xxxx y x2 5x4y x2 3a-2 b5 = 3 bbbbb bbb = b83a-2 b5 = a(-2-4)b(5-(-3)) = a-6 b8 = b8 9a4b-3 9aaaa aa 3a6 9a4b-3 3 3 3a6

Powers to Powers (am)n = amn (a2)3 a2• a2 • a2 = aa aa aa = a6 (24)-2 = 1 = 1 = 1 = 1/256 (24)2 24• 24 16 • 16 28 256 (x3)-2 = x –6 = x 10 = x4 (x -5)2 x –10 x 6 (24)-2 = 2-8 = 1 = 1

Products to Powers (ab)n = anbn (6y)2 = 62y2 = 36y2 (2a2b-3)2 = 22a4b-6 = 4a4 = a 4(ab3)3 4a3b9 4a3b9b6 b15 What about this problem? 5.2 x 1014 = 5.2/3.8 x 109 1.37 x 109 3.8 x 105 Do you know how to do exponents on the calculator?

Square Roots & Cube Roots A number b is a cube root of a number a if b3 = a 8 = 2 since 23 = 8 Notice that 8 breaks down into 2 • 2 • 2 So, 8 =  2 • 2 • 2 See a ‘group of 3’ –> bring it outside the radical (the cube root sign) Example: 200 = 2 • 100 = 2 • 10 • 10 = 2 • 5 • 2 • 5 • 2 = 2 • 2 • 2 • 5 • 5 = 2 25 A number b is a square root of a number a if b2 = a 25 = 5 since 52 = 25 Notice that 25 breaks down into 5 • 5 So, 25 =  5 • 5 See a ‘group of 2’ -> bring it outside the radical (square root sign). Example: 200 = 2 • 100 = 2 • 10 • 10 = 10 2 3 3 3 3 3 3 3 3 Note: -25 is not a real number since no number multiplied by itself will be negative Note: -8 IS a real number (-2) since -2 • -2 • -2 = -8 3

     Nth Root ‘Sign’ Examples Even radicals of positive numbers Have 2 roots. The principal root Is positive. 16 = 4 or -4 not a real number -16 Even radicals of negative numbers Are not real numbers. 4 -16 not a real number Odd radicals of negative numbers Have 1 negative root. 5 -32 = -2 5 Odd radicals of positive numbers Have 1 positive root. 32 = 2

m Xm Ym X Y = Exponent Rules (XY)m = xmym

Examples to Work through

Some Rules for Simplifying Radical Expressions

Practice Problems

Operations on Radical Expressions • Addition and Subtraction (Combining LIKE Terms) • Multiplication and Division • Rationalizing the Denominator

Radical Operations with Numbers

Multiplying Radicals (FOIL works with Radicals Too!)

Rationalizing the Denominator • Remove all radicals from the denominator

Adding & Subtracting Polynomials Combine Like Terms (2x2 –3x +7) + (3x2 + 4x – 2) = 5x2 + x + 5 (5x2 –6x + 1) – (-5x2 + 3x – 5) = (5x2 –6x + 1) + (5x2 - 3x + 5) = 10x2 – 9x + 6 Types of Polynomials f(x) = 3 Degree 0 Constant Function f(x) = 5x –3 Degree 1 Linear f(x) = x2 –2x –1 Degree 2 Quadratic f(x) = 3x3 + 2x2 – 6 Degree 3 Cubic

Multiplication of Polynomials Step 1: Using the distributive property, multiply every term in the 1st polynomial by every term in the 2nd polynomial Step 2: Combine Like Terms Step 3: Place in Decreasing Order of Exponent 4x2 (2x3 + 10x2 – 2x – 5) = 8x5 + 40x4 –8x3 –20x2 (x + 5) (2x3 + 10x2 – 2x – 5) = 2x4 + 10x3 – 2x2 – 5x + 10x3 + 50x2 – 10x – 25 = 2x4 + 20x3 + 48x2 –15x -25

Binomial Multiplication with FOIL (2x + 3) (x - 7) F. O. I. L. (First) (Outside) (Inside) (Last) (2x)(x) (2x)(-7) (3)(x) (3)(-7) 2x2 -14x 3x -21 2x2 -14x + 3x -21 2x2 - 11x -21

Division by a Monomial 3x2 + x 5x3 – 15x2 x 15x 4x2 + 8x – 12 5x2y + 10xy2 4x2 5xy 15A2 – 8A2 + 12 12A5 – 8A2 + 12 4A 4A

Review: Factoring Polynomials To factor a number such as 10, find out ‘what times what’ = 10 10 = 5(2) To factor a polynomial, follow a similar process. Factor: 3x4 – 9x3 +12x2 3x2 (x2 – 3x + 4) Another Example: Factor 2x(x + 1) + 3 (x + 1) (x + 1)(2x + 3)

Solving Polynomial Equations By Factoring Zero Product Property : If AB = 0 then A = 0 or B = 0 Solve the Equation: 2x2 + x = 0 Step 1: Factor x (2x + 1) = 0 Step 2: Zero Product x = 0 or 2x + 1 = 0 Step 3: Solve for X x = 0 or x = - ½ Question: Why are there 2 values for x???

Factoring Trinomials To factor a trinomial means to find 2 binomials whose product gives you the trinomial back again. Consider the expression: x2 – 7x + 10 (x – 5) (x – 2) The factored form is: Using FOIL, you can multiply the 2 binomials and see that the product gives you the original trinomial expression. How to find the factors of a trinomial: Step 1: Write down 2 parentheses pairs. Step 2: Do the FIRSTS Step3 : Do the SIGNS Step4: Generate factor pairs for LASTS Step5: Use trial and error and check with FOIL

Practice • Factor: • y2 + 7y –30 4. –15a2 –70a + 120 • 10x2 +3x –18 5. 3m4 + 6m3 –27m2 • 8k2 + 34k +35 6. x2 + 10x + 25

Special Types of Factoring Square Minus a Square A2 – B2 = (A + B) (A – B) Cube minus Cube and Cube plus a Cube (A3 – B3) = (A – B) (A2 + AB + B2) (A3 + B3) = (A + B) (A2 - AB + B2) Perfect Squares A2 + 2AB + B2 = (A + B)2 A2 – 2AB + B2 = (A – B)2

Quadratic Equations • General Form of Quadratic Equation • ax2 + bx + c = 0 • a, b, c are real numbers & a 0 • A quadratic Equation: x2 – 7x + 10 = 0 a = _____ b = _____ c = ______ • Methods & Tools for Solving Quadratic Equations • Factor • Apply zero product principle (If AB = 0 then A = 0 or B = 0) • Square root method • Completing the Square • Quadratic Formula 1 -7 10 Example1:Example 2: x2 – 7x + 10 = 0 4x2 – 2x = 0 (x – 5) (x – 2) = 0 2x (2x –1) = 0 x – 5 = 0 or x – 2 = 0 2x=0 or 2x-1=0 + 5 + 5 + 2 + 2 2 2 +1 +1 2x=1 x = 5 or x = 2 x = 0 or x=1/2

Square Root Method If u2 = d then u = d or u = - d. If u2 = d then u = + d • Solving a Quadratic Equation with the Square Root Method • Example 1: Example 2: • 4x2 = 20 (x – 2)2 = 6 • 4 • x – 2 = +6 • x2 = 5 + 2 + 2 • x = + 5 x = 2 + 6 • So, x = 5 or - 5 So, x = 2 + 6 or 2 - 6

Completing the Square (Example 1) If x2 + bx is a binomial then by adding b2 which is the square of half 2 the coefficient of x, a perfect square trinomial results: x2 + bx + b2 = x + b2 2 2 Solving a quadratic equation with ‘completing the square’ method. Example: Step1: Isolate the Binomial x2 - 6x + 2 = 0 -2 -2 Step 2: Find ½ the coefficient of x (-3 ) x2 - 6x = -2 and square it (9) & add to both sides. x2 - 6x + 9 = -2 + 9 (x – 3)2 = 7 x – 3 = + 7 x = (3 + 7 ) or (3 - 7 ) Note: If the coefficient of x2 is not 1 you must divide by the coefficient of x2 before completing the square. ex: 3x2 – 2x –4 = 0 (Must divide by 3 before taking ½ coefficient of x) Step 3: Apply square root method

(Completing the Square – Example 2) Step 1: Check the coefficient of the x2 term. If 1 goto step 2 If not 1, divide both sides by the coefficient of the x2 term. Step 2: Calculate the value of : (b/2)2 [In this example: (2/2)2 = (1)2 = 1] Step 3: Isolate the binomial by grouping the x2 and x term together, then add (b/2)2 to both sides of he equation. Step 4: Factor & apply square root method 2x2 +4x – 1 = 0 2x2 +4x – 1 = 0 (x + 1) (x + 1) = 3/2 2 2 2 2 (x + 1)2 = 3/2 x2 +2x – 1/2 = 0 (x2 +2x ) = ½ √(x + 1)2 = √3/2 (x2 +2x + 1 ) = 1/2 + 1 x + 1 = +/- √6/2 x = √6/2 – 1 or - √6/2 - 1

Quadratic Formula General Form of Quadratic Equation: ax2 + bx + c = 0 Quadratic Formula: x = -b + b2 – 4ac discriminant: b2 – 4ac 2a if 0, one real solution if >0, two unequal real solutions if <0, imaginary solutions Solving a quadratic equation with the ‘Quadratic Formula’ 2x2–6x + 1= 0 a = ______ b = ______ c = _______ x = - (-6) + (-6)2 – 4(2)(1) 2(2) = 6 + 36 –8 4 = 6 + 28 = 6 + 27 = 2 (3 + 7 ) = (3 + 7 ) 4 4 4 2 2 -6 1

Solving Higher Degree Equations x3 = 4x x3 - 4x = 0 x (x2 – 4) = 0 x (x – 2)(x + 2) = 0 x = 0 x – 2 = 0 x + 2 = 0 x = 2 x = -2 2x3 + 2x2 - 12x = 0 2x (x2 + x – 6) = 0 2x (x + 3) (x – 2) = 0 2x = 0 or x + 3 = 0 or x – 2 = 0 x = 0 or x = -3 or x = 2

Solving By Grouping x3 – 5x2 – x + 5 = 0 (x3 – 5x2) + (-x + 5) = 0 x2 (x – 5) – 1 (x – 5) = 0 (x – 5)(x2 – 1) = 0 (x – 5)(x – 1) (x + 1) = 0 x – 5 = 0 or x - 1 = 0 or x + 1 = 0 x = 5 or x = 1 or x = -1

Rational Expressions Rational Expression – an expression in which a polynomial is divided by another nonzero polynomial. • Examples of rational expressions • x 2 • x 2x – 5 x – 5 • Domain = {x | x  0} Domain = {x | x  5/2} Domain = {x | x  5}

Multiplication and Division of Rational Expressions A • C = A 9x = 3 B • C B 3x2 x 5y – 10 = 5 (y – 2) = 5 = 1 10y - 20 10 (y – 2) 10 2 2z2 – 3z – 9 = (2z + 3) (z – 3) = 2z + 3 z2 + 2z – 15 (z + 5) (z – 3) z + 5 A2 – B2 = (A + B)(A – B) = (A – B) A + B (A + B)

Negation/Multiplying by –1 -y – 2 4y + 8 y + 2 4y + 8 -y - 2 -4y - 8 - = OR

Examples x3 – x x + 1 x – 1 x x2 – 25 x2 –10x + 25 x2 + 5x + 4 2x2 + 8x •  x2 – 25 2x2 + 8x x2 + 5x + 4 x2 –10x + 25 • (x3 – x) (x + 1) x(x – 1) = = (x + 5) (x – 5) • 2x(x + 4) (x + 4)(x + 1) • (x – 5) (x – 5) = = x (x2 – 1)(x + 1) x(x – 1) 2x (x + 5) (x + 1)(x – 5) = x (x + 1) (x – 1)(x + 1) x(x – 1) = (x + 1)(x + 1) = (x + 1)2 =

(x + 1) (x –7) (x + 1) (x – 1) 1 x2 + x - 6 x – 2 3 • 1 (x + 3) (x – 2) x – 2 3 (x – 7) (x – 1) • (x + 3) 3 Check Your Understanding Simplify: x2 –6x –7 x2 -1 Simplify: 1 3 x - 2 x2 + x - 6 

Addition of Rational Expressions Adding rational expressions is like adding fractions With LIKE denominators: 1 + 2 = 3 8 8 8 x + 3x - 1 = 4x - 1 x + 2 x + 2 x + 2 x + 2 (2 + x) (2 + x) 3x2 + 4x - 4 3x2 + 4x -4 (3x2 + 4x – 4) (3x -2)(x + 2) = = = 1 (3x – 2)

Adding with UN-Like Denominators • + 2 • x2 – 9 x + 3 • 1 + 2 • (x + 3)(x – 3) (x + 3) • 1 + 2 (x – 3) • (x + 3)(x – 3) (x + 3)(x – 3) • 1 + 2(x – 3) 1 + 2x – 6 2x - 5 • (x + 3) (x – 3) (x + 3) (x – 3) (x + 3) (x – 3) • + 1 • 8 • (3) (2) + 1 • 8 • + 1 • 8 • 7 • 8 = =

x – 1 (x + 1)(x –1) = = 1 (x + 1) Subtraction of Rational Expressions To subtract rational expressions: Step 1: Get a Common Denominator Step 2: Combine Fractions DISTRIBUTING the ‘negative sign’ BE CAREFUL!! 2x - x + 1 x2 – 1 x2 - 1 2x – (x + 1) x2 -1 2x – x - 1 x2 -1 = =

b b-1 2(b – 2) b-2 - b -b+1 2(b – 2) b-2 + b 2(b – 2) 2(-b+1) 2(b – 2) + -1 2 -1(b – 2) 2(b – 2) b –2b+2 2(b – 2) -b + 2 2(b – 2) = = = Check Your Understanding Simplify: b b-1 2b - 4 b-2 -

x + 2 3x - 1 x x + 4 Complex Fractions A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both. Examples: 1 5 4 7 x x2 – 16 1 x - 4 1 x 2 x2 + 3 x 1 x2 - 7/20

(2x – 1) (x - 2) (x + 1) Rational Equations 3x = 3 x + 1 = 3 6 = x 2x – 1 x – 2 x - 2 x + 1 3x = 3(2x – 1) 3x = 6x – 3 -3x = -3 x = 1 x + 1 = 3 x = 2 6 = x (x + 1) 6 = x2 + x x2 + x – 6 = 0 (x + 3 ) (x - 2 ) = 0 x = -3 or x = 2 Careful! – What do You notice about the answer?

(12x) 6 (x + 1) -3(x – 1) = 4x 6x + 6 –3x + 3 = 4x 3x + 9 = 4x -3x -3x 9 = x = Rational Equations Cont… To solve a rational equation: Step 1: Factor all polynomials Step 2: Find the common denominator Step 3: Multiply all terms by the common denominator Step 4: Solve x + 1 - x – 1 = 1 2x 4x 3

(4x2) (x + 2)(x – 2) 4x + 4 = 3x2 3x2 - 4x - 4 = 0 (3x + 2) (x – 2) = 0 3x + 2 = 0 or x – 2 = 0 3x = -2 or x = 2 x = -2/3 or x = 2 3(x + 2) + 5(x – 2) = 12 3x + 6 + 5x – 10 = 12 8x – 4 = 12 + 4 + 4 8x = 16 x = 2 Other Rational Equation Examples 3 + 5 = 12 x – 2 x + 2 x2 - 4 1 + 1 = 3 x x2 4 3 + 5 = 12 x – 2 x + 2 (x + 2) (x – 2)

Solve for p: • = 1 + 1 • F p q 1 x - 1 2(x – 3) x(x – 2) 3 x(x – 1)(x + 1) Check Your Understanding Simplify: x 1 x2 – 1 x2 – 1 1 3 x – 2 x 1 1 2 x(x – 1) x2 – 1 x(x + 1) Solve 6 1 x 2 3 2 2x – 1 x + 1 2 3 x x – 1 x + 2 x2 + x - 2 4 + - = 1 = 5 - + = + - -1/4 Try this one:

Radical Equations Continued… Example1: x + 26 – 11x = 4 26 – 11x = 4 - x (26 – 11x)2 = (4 – x)2 26 – 11x = (4-x) (4-x) 26 - 11x = 16 –4x –4x +x2 26 –11x = 16 –8x + x2 -26 +11x -26 +11x 0 = x2 + 3x -10 0 = (x - 2) (x + 5) x – 2 = 0 or x + 5 = 0 x = 2 x = -5 Example 2: X2= 64 Example 3:

Inequality Set & Interval Notation Set Builder Notation {1,5,6} { }  {6} {x | x > -4} {x | x < 2} {x | -2 < x < 7} x such that x such that x is less x such that x is greater x is greater than –4 than or equal to 2 than –2 and less than or equal to 7 Interval (-4, ) (-, 2] (-2, 7] Notation Graph -4 0 2 7 -2 Question: How would you write the set of all real numbers? (-, ) or R

Inequality Example StatementReason 7x + 15 > 13x + 51 [Given] -6x + 15 > 51 [-13x] -6x > 36 [-15] x < 6 [Divide by –6, so must ‘flip’ the inequality sign Set Notation: {x | x < 6} Interval Notation: (-, 6] Graph: 6

Review Topics (Chapter 0 & 1)