b) 24h2 20h 24 – – for common factors; 4 is a common factor. So, 24h2 20h 24 = 4(6h2 5h 6) – – – – Factor 6h2 5h 6. – – The product of the coefficient of and the constant term is: 6(–6) = h2 –36 Write as the sum of two terms whose coefficients have a product of –5h –36. Factors of Sum of Factors –36 1, 1 36 = –36 – –35 36 + 36 = 35 –1, –1 2, 2 18 = –18 – –16 18 + 18 = 16 –2, –2 3, 3 12 = –12 – –9 12 + 12 = 9 –3, –3 4, 4 9 = –9 – –5 9 + 9 = 5 –4, –4 6, 6 6 = 0 –6 – 6 + 6 = 0 –6, –6 The two coefficients are 4 and so write the trinomial –9, 6h2 5h 6 as 6h2 + 4h 9h 6. – – – – Remove a common factor from the 1st pair of terms, and from the 2nd pair of terms: 6h2 + 4h 9h 6 = 2h(3h + 2) 3(3h + 2) – – – Each product has a common binomial factor. 6h2 + 4h 9h 6 = (3h + 2)(2h 3) – – – So, 6h2 5h 6 = (3h + 2)(2h 3) – – – And, 24h2 20h 24 = 4(3h + 2)(2h 3) – – – Homework #6 ignore tiles #8-10, 12-19 #21
b) 24h2 – 20h – 24 for common factors; 4 is a common factor. So, 24h2 – 20h – 24 = 4(6h2 – 5h – 6) Factor 6h2 – 5h – 6. The product of the coefficient of h2 and the constant term is: 6(–6) = –36 Write –5h as the sum of two terms whose coefficients have a product of –36. Factors of –36 Sum of Factors 1, –36 1 – 36 = –35 –1, 36 –1 + 36 = 35 2, –18 2 – 18 = –16 –2, 18 –2 + 18 = 16 3, –12 3 – 12 = –9 –3, 12 –3 + 12 = 9 4, –9 4 – 9 = –5 –4, 9 –4 + 9 = 5 6, –6 6 – 6 = 0 –6, 6 –6 + 6 = 0 The two coefficients are 4 and –9, so write the trinomial 6h2 – 5h – 6 as 6h2 + 4h – 9h – 6. Remove a common factor from the 1st pair of terms, and from the 2nd pair of terms: 6h2 + 4h – 9h – 6 = 2h(3h + 2) – 3(3h + 2) Each product has a common binomial factor. 6h2 + 4h – 9h – 6 = (3h + 2)(2h – 3) So, 6h2 – 5h – 6 = (3h + 2)(2h – 3) And, 24h2 – 20h – 24 = 4(3h + 2)(2h – 3) Homework #6 ignore tiles #8-10, 12-19 #21
miap_ch3_notes.docx