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

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Presentation on theme: "Math 20-1 Chapter 4 Quadratic Equations 3.1 Factoring Quadratic Equations Teacher Notes."— Presentation transcript:

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

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

3 4.2A.4

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

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

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

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

8 Factor Factors of A.7

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

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

11 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) Suggested Questions: Part A p PartB (7-10)ac,11a,12a 13b 14 15a 16b 18c 19 mc 4.2A.10

12 Dealing with Fractions in Factoring

13 4.2A Factoring Quadratic Expressions 4.2A.2 Factor Greatest Common Factor Simple Trinomial Perfect Square Trinomial Differenc e of Squares Decomp osition Fractions Complex Bases


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