# Math 20-1 Chapter 4 Quadratic Equations

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Math 20-1 Chapter 4 Quadratic Equations
Teacher Notes 3.1 Factoring Quadratic Equations

x2 + (a + b)x + ab = (x + a)(x + b)
Solve by Factoring Simple Trinomials Simple Trinomials: The coefficient of the x2 term is 1. Recall: (x + 6)(x + 4) = x(x + 4) + 6(x + 4) = x2 + 4x + 6x + (6)(4) = x2 + (4 + 6)x + (6)(4) = x2 + 10x + 24 The last term is the product of the constant terms. Explain each step going backwards The middle term is the sum of the constant terms of the binomial. x2 + (a + b)x + ab = (x + a)(x + b) Factor: x2 + 9x + 20 Factors of 20 1 20 2 10 4 5 x2 + (5 + 4)x + (4)(5) x2 + 5x + 4x + (4)(5) x(x + 5) + 4(x + 5) (x + 5)(x + 4) 4.2A.3

Factors of 12 1 12 2 6 3 4 4.2A.4

Factoring General Trinomials Decomposition
Recall: (3x + 2)(x + 5) = 3x(x + 5) +2(x + 5) = 3x2 + 15x + 2x + 10 = 3x2 + 17x + 10 Explain each step going backwards To factor 3x2 + 17x + 10, find two numbers that have a product of 30 and a sum of 17. Factors of 30 1 30 2 15 3 10 4.2A.5

Solving General Trinomials - the Decomposition Method
Factors of 24 1 24 2 12 3 8 4 6 3x2 - 10x + 8 The product is 3 x 8 = 24. The sum is -10. 3x2 - 6x - 4x + 8 Rewrite the middle term of the polynomial using -6 and -4. (-6x - 4x is just another way of expressing -10x.) 3x(x - 2) - 4(x - 2) Factor by grouping. (x - 2)(3x - 4) 4.2A.6

Factoring General Trinomials - Some Examples
1. 6x2 + x - 15 product is - 90 sum is 1 10 & -9 = 6x2 + 10x - 9x - 15 = 2x(3x + 5) - 3(3x + 5) Check using foil: (3x + 5)(2x - 3) = 6x2 - 9x + 10x + 15 = 6x2 + x - 15 = (3x + 5)(2x - 3) y2 - 5y - 3 product is -36 sum is -5 4 & -9 = 12y2 - 9y + 4y - 3 = 3y(4y - 3) + 1(4y - 3) = (4y - 3)(3y + 1) 3. 3x2 + 20x + 12 product is 36 sum is 20 2 & 18 = 3x2 + 18x + 2x + 12 = 3x(x + 6) + 2(x + 6) = (x + 6)(3x + 2)

Factoring General Trinomials - the Inspection Method
1. List the pairs of factors of the first and last terms. Factor: x2 - 11x + 3 2 x 1= 2 3 x 3 = 9 Sum = 11 1 x 1 = 1 6 x 3 = 18 Sum = 19 1 x 3 = 3 6 x 1 = 6 Sum = 9 2. Do a cross product sum to find the sum equal to the middle term. 1 6 2 3 1 3 3. Place the terms in the brackets, keeping in mind FOIL. (place in opposite brackets.) Not equal to the middle term so don’t use. Try again. This sum is equal to the middle term, so use these numbers. Not equal to the middle term so don’t use. Try again. (2x - 3)(3x - 1) Check using foil: 6x2 - 2x - 9x + 3 = 6x2 - 11x + 3

Factor Factors of 4 1 4 2 4.2A.7

Recall that, when multiplying conjugate binomials, the
Difference of Squares Recall that, when multiplying conjugate binomials, the product is a difference of squares. (x - 7)(x + 7) = x2 + 7x – 7x - 49 = x2 - 49 Therefore, when factoring a difference of squares, the factors will be conjugate binomials. Factor: x = (x - 9)(x + 9) 16x = (4x - 11)(4x + 11) (x)2 - (9)2 (4x)2 - (11)2 5x2 - 80y2 = 5(x2 - 16y2) = 5(x - 4y)(x + 4y) 4.2A.8

Factoring a Difference of Squares with a Complex Base
Factor completely: (x + y) = B = [B - 4] [B + 4] = [(x + y) - 4] [(x + y) + 4] = (x + y - 4)(x + y + 4) 25 - (x + 3)2 = [5 - (x + 3)] [5 + (x + 3)] = (5 - x - 3)(5 + x + 3) = (-x + 2)(8 + x) 4.2A.9

25(x - 1)2 - 4(3x + 2)2 = [5(x - 1) - 2(3x + 2)] [5(x - 1) + 2(3x + 2)] = (5x x - 4)(5x x + 4) = (-x - 9)(11x - 1) Assignment Suggested Questions: Part A p PartB (7-10)ac,11a,12a 13b 14 15a 16b 18c 19 mc 4.2A.10

Dealing with Fractions in Factoring

Greatest Common Factor Simple Trinomial Perfect Square Trinomial
4.2A Factoring Quadratic Expressions Factor Greatest Common Factor Simple Trinomial Perfect Square Trinomial Difference of Squares Decomposition Fractions Complex Bases 4.2A.2

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