Lesson 3.5 Check Your Understanding Outcome A5 page 159-165 Expand and simplify. a) (c + 3)(c – 7) b) (5 – s)(9 – s) Solutions a) (c + 3)(c – 7) Method 1 Use a rectangle diagram. (c + 3)(c – 7) = c2 + (–7c) + 3c + (–21) Combine like terms. = c2 – 4c – 21 Method 2 Use the distributive property. (c + 3)(c – 7) = c(c – 7) + 3(c – 7) = (c)(c) + (c)(–7) + (3)(c) + (3)(–7) = c2 – 7c + 3c – 21 Combine like terms. = c2 – 4c – 21 b) (5 – s)(9 – s) Use the distributive property. (5 – s)(9 – s) = 5(9 – s) + (–s)(9 – s) = (5)(9) + (5)(–s) + (–s)(9) + (–s)(–s) = 45 – 5s – 9s + s2 = 45 – 14s + s2
2. Factor each trinomial. a) x2 – 8x + 7 b) a2 + 7a – 18 Solution a) x2 – 8x + 7 Since factoring and expanding are inverse processes, there will be two binomial factors. They will have the form: (x + an integer)(x + an integer) The two integers in the binomial have a sum of –8 and a product of 7. To determine the integers, list of factors of 7, then add the factors in each pair. Since the product is positive, the factors have the same sign. Factors of 7 Sum of Factors –1, –7 –1 – 7 = –8 1, 7 1 + 7 = 8 The factors that have a sum of –8 are –1 and –7. Write these integers as the second terms in the binomials. x2 – 8x + 7 = (x – 1)(x – 7) b) a2 + 7a – 18 Find two integers whose sum is 7 and whose product is –18. Since the product is negative, the factors have different signs. Factors of –18 Sum of Factors –1, 18 –1 + 18 = 17 1, –18 1 – 18 = –17 –2, 9 –2 + 9 = 7 2, –9 2 – 9 = –7 –3, 6 –3 + 6 = 3 3, –6 3 – 6 = –3 The factors that have a sum of 7 are –2 and 9. a2 + 7a – 18 = (a – 2)(a + 9)
3. Factor: –30 + 7m + m2 Solutions Method 1 –30 + 7m + m2 The binomials will have the form: (an integer + m)(an integer + m) Find two integers whose product is –30 and whose sum is 7. The integers are –3 and 10. So, –30 + 7m + m2 = (–3 + m)(10 + m) Method 2 Rewrite the polynomial in descending order. –30 + 7m + m2 = m2 + 7m – 30 Find two integers whose product is –30 and whose sum is 7. The integers are –3 and 10. m2 + 7m – 30 = (m – 3)(m + 10)
4. Factor. –5h2 – 20h + 60 Solution –5h2 – 20h + 60 The greatest common factor is 5. Since the coefficient of the first term is negative, use –5 as the common factor. –5h2 – 20h + 60 = –5(h2 + 4h – 12) Two numbers with a sum of 4 and a product of –12 are –2 and 6. So, h2 + 4h – 12 = (h – 2)(h + 6) And, –5h2 – 20h + 60 = –5(h – 2)(h + 6) Since factoring and expanding are inverse processes, check that the factors are correct. Multiply the factors. –5(h – 2)(h + 6) = –5(h2 + 4h – 12) = –5h2 – 20h + 60 This trinomial is the same as the original trinomial, so the factors are correct. Homework read page 159-165 Do page 166 #7 #9, 10, 11, 12, 14, 15, 21 #23 Ignore tiles and sketches
Method 2 4g2 + 11g + 6 Use number sense and reasoning, with mental math. The factors of 4g2 are 4g and 1g, or 2g and 2g. The factors of 6 are 6 and 1, or 3 and 2. Arrange the factor combinations vertically. 4g 6 2g 6 Form the products, then add. 1g 1 2g 1 4g + 6g = 10g 2g + 12g = 14g These combinations don’t work. Try a different combination. 4g 3 Form the products, then add. 1g 2 8g + 3g = 11g This is the correct combination of factors. So, 4g2 + 11g + 6 = (4g + 3)(g + 2) b) Method 1 6m2 – 7m – 10 Find the factors of the form (am + b)(cm + d). The coefficient of m2 is 6, so the coefficients of the 1st terms in the binomial are factors of 6, which are 1 and 6, or 2 and 3. The constant term in the trinomial is –10, so the 2nd in the binomial are factors of –10, which are 1 and –10, or –1 and 10, or –2 and 5, or 2 and –5. The coefficient of the middle term of the trinomial, –7, is between the coefficients of the other two terms, so try combinations of the lesser factors first; that is, try combinations of 1 and 6 with combinations of ±2 and ±5. So, the binomials could be: (m + 2)(6m – 5) or (m – 5)(6m + 2) or (m – 2)(6m + 5) or (m + 5)(6m – 2) Check if any of the 4 binomial products above has its m-term equal to –7m. (m + 2)(6m – 5) = 6m2 + 7m – 10 (m – 5)(6m + 2) = 6m2 – 28m – 10 (m – 2)(6m + 5) = 6m2 – 7m – 10 This is the correct trinomial so we need not any further. 6m2 – 7m – 10 = (m – 2)(6m + 5) (Solution continues.)
4. Factor. a) 8p2 – 18p – 5 b) 24h2 – 20h – 24 Solution a) 8p2 – 18p – 5 Check for common factors; there are none. The product of the coefficient of p2 and the constant term is: 8(–5) = –40 Write –18p as the sum of two terms whose coefficients have a product of –40. Factors of –40 Sum of Factors 1, –40 1 – 40 = –39 –1, 40 –1 + 40 = 39 2, –20 2 – 20 = –18 –2, 20 –2 + 20 = 18 4, –10 4 – 10 = –6 –4, 10 –4 + 10 = 6 5, –8 5 – 8 = –3 –5, 8 –5 + 8 = 3 The two coefficients are 2 and –20, so write the trinomial 8p2 – 18p – 5 as 8p2 + 2p – 20p – 5. Remove a common factor from the 1st pair of terms, and from the 2nd pair of terms: 8p2 + 2p – 20p – 5 = 2p(4p + 1) – 5(4p + 1) Each product has a common binomial factor. 8p2 + 2p – 20p – 5 = (4p + 1)(2p – 5) So, 8p2 – 18p – 5 = (4p + 1)(2p – 5) (Solution continues.)
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
3. Expand and simplify. a) (4m + 1)(3m – 2) + 2(2m – 1)(–3m + 4) b) (6h + k – 2)(2h – 3) – (4h – 3k)2 Solution Use the order of operations. Multiply before adding and subtracting. Then combine like terms. a) (4m + 1)(3m – 2) + 2(2m – 1)(–3m + 4) = 4m(3m – 2) + 1(3m – 2) + 2[(2m(–3m + 4) – 1(–3m + 4)] = 12m2 – 8m + 3m – 2 + 2[–6m2 + 8m + 3m – 4] = 12m2 – 5m – 2 + 2[–6m2 + 11m – 4] = 12m2 – 5m – 2 – 12m2 + 22m – 8 = 17m – 10 b) (6h + k – 2)(2h – 3) – (4h – 3k)2 = 6h(2h – 3) + k(2h – 3) – 2(2h – 3) – (4h – 3k)(4h – 3k) = 12h2 – 18h + 2hk – 3k – 4h + 6 – [4h(4h – 3k) – 3k(4h – 3k)] = 12h2 – 22h + 2hk – 3k + 6 – [16h2 – 12hk – 12hk + 9k2] = 12h2 – 22h + 2hk – 3k + 6 – (16h2 – 24hk + 9k2) = 12h2 – 22h + 2hk – 3k + 6 – 16h2 + 24hk – 9k2 = –4h2 – 22h + 26hk – 3k + 6 – 9k2 Homework page 186 #4-5 #6a, 7, 9-12, 14-15, 17 #21-22