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

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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)

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4.2A.4

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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

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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

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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

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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

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Factor Factors of A.7

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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

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= [(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

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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

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Dealing with Fractions in Factoring

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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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