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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. 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 a2 + b2 = c2 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.