S1A9 Factoring Simple Trinomials & Special Products PDF

S1A9: Factoring Simple Trinomials and Special Products (challenge option in red) Factor each expression into a product. If not factorable, label as prime. 2 2 2 1) 𝑥 + 5𝑥 + 4...

S1A9: Factoring Simple Trinomials and Special Products (challenge option in red) Factor each expression into a product. If not factorable, label as prime. 2 2 2 1) 𝑥 + 5𝑥 + 4 2) 𝑞 + 16𝑞 + 15 3) 42 − 23𝑘 + 𝑘 2 2 2 3 2 4) 𝑥 − 16𝑥𝑦 + 45𝑦 5) (𝑥 + 2) + 4(𝑥 + 2) − 21 6) 𝑥 + 4𝑥 − 12𝑥 Challenge: Find all the values for 𝑘 for which the expression is factorable. 2 2 1) 𝑦 + 𝑘𝑦 + 14 2) 𝑧 + 𝑘𝑧 + 12 Find all the positive values for 𝑘 for which the expression is factorable. (#3-4) 2 2 3) 𝑛 + 6𝑛 + 𝑘 4) 𝑦 + 8𝑦 + 𝑘 Factor each expression into a product given 𝑛 is a positive integer. (#5-6) 2𝑛 𝑛 2𝑛 4𝑛 4𝑛 2𝑛 𝑛 2𝑛 5) 𝑎 − 30𝑎 𝑏 + 209𝑏 6) 𝑝 − 30𝑝 𝑞 + 221𝑞 2 2 7) Factor (𝑎 + 𝑏) − 𝑐(𝑎 + 𝑏) − 2𝑐 Factor each expression into a product. Use the “Zero Product Property” to solve. 2 2 7) 3𝑥(2𝑥 + 1)(2𝑥 + 5) = 0 8) 0 = 𝑘 − 12𝑘 + 35 9) 𝑚 − 36 = 16𝑚 Challenge: Find the equation in standard form ( 2 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0, 𝑤ℎ𝑒𝑟𝑒 𝑎 𝑏 & 𝑐 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠) for which the solutions are: 8) {2, − 3} 9) { 5 2 ,− 2 } Factor each expression into a product. If not factorable, label as prime. 2 2 2 10) 𝑏 − 36 11) 49𝑎 − 9𝑏 5 2 6 2 2 12) 32𝑥 𝑦 − 2𝑥𝑦 13) (𝑥 + 4) −𝑥 2 2 2 14) 𝑥 + 6𝑥 + 9 15) 4𝑠 − 36𝑠𝑡 + 81𝑡 Factor each expression into a product. Use the “Zero Product Property” to solve. 2 2 3 16) 𝑟 − 81 = 0 17) 288𝑦 + 48𝑦 + 2𝑦 =0 Simplify and express in descending order (of powers) for the given variable: 1) (3𝑥 − 2𝑦 + 5) + (𝑥 + 2𝑦 − 1) 2) (5𝑟 − 2𝑦) − (2𝑟 − 3𝑦) 3 2 3 2 3 ( 3) 3𝑎 − 2𝑎𝑏 ) − (𝑎 − 4𝑎𝑏 −𝑏 ) 4) (𝑦 + 3)(𝑦 + 2) 5) (𝑎 + 4)(𝑎 − 1) 6) (2𝑥 − 1)(𝑥 − 5) 2 2 7) (𝑎 + 2)(𝑎 + 3𝑎 + 5) 8) (3𝑧 − 2)(3𝑧 − 𝑧 + 4) 2 9) The square of 𝑥 − 3𝑥 + 5 10) The cube of 𝑥 + 5 Factor each expression by factoring out the GCF(greatest common factor) 2 11) 15𝑎 − 25𝑏 + 20 12) 6𝑥 + 10𝑥 2 3 2 2 13) 6𝑝 𝑞 − 9𝑝𝑞 14) 48𝑎 𝑏 + 72𝑎 𝑏 15) 3(𝑥 + 𝑦) + 𝑧(𝑥 + 𝑦) 16) 𝑒(𝑓 − 𝑔) + 4(𝑔 − 𝑓) Solve each factored expression using the “Zero Product Property.” 17) (𝑦 + 5)(𝑦 − 7) = 0 18) 15𝑛(𝑛 + 15) = 0 19) (2𝑡 − 3)(3𝑡 − 2) = 0 20) 3𝑥(2𝑥 + 1)(2𝑥 + 5) = 0 Solve each equation using the “Zero Product Property” or the “Square Root Principle.” 2 3 2 21) 𝑦 = 16𝑦 22) 8𝑢 − 2𝑢 =0 2 23) (𝑥 − 2) − 6 = 19 24) 𝑥(𝑥 + 1) − 2(𝑥 + 1) = 0 Factor each expression by “grouping” 25) 3𝑎 + 𝑎𝑏 + 3𝑐 + 𝑏𝑐 26) 3ℎ𝑘 − 2𝑘 − 12ℎ + 8 2 2 2 2 3 2 27) (ℎ 𝑘 + 4𝑘 ) + (ℎ 𝑘 + 4𝑘) 28) 𝑥 − 3𝑥 −𝑥+3 Part 1 Choose at least 4. Factor the expression into a product. If not factorable, label as prime. 2 2 2 1) 3𝑥 + 7𝑥 + 2 2) 5𝑦 + 4𝑦 − 1 3) 7𝑥 + 8𝑥 + 1 2 2 2 2 4) 5 + 7𝑥 − 6𝑥 5) 6𝑟 − 11𝑟𝑝 + 5𝑝 6) 2𝑥 + 14𝑥 + 3 2 2 2 7) 16𝑎 − 81 8) 𝑟 − 5𝑟 − 14 9) 16𝑦 + 24𝑦 + 9 Part 1 Choose at least 4. Factor each expression into a product. If not factorable, label as prime. 2 2 2 1) 2(𝑎 + 2) + 5(𝑎 + 2) − 3 2) (𝑦 + 3𝑦 − 1) − 9  4 2 2 2 3) 4𝑥 − 17𝑥 + 4 4) 72(𝑎 + 2𝑏) + 5(𝑎 + 2𝑏)𝑐 − 3𝑐 2 2 4 4 2 5) 9𝑥 − 12𝑥 + 4 − 81 6) 4𝑥 𝑦 − 36𝑥 𝑦 2 2 2 7) Show that (15𝑥 − 14𝑥 + 3)(6𝑥 + 19𝑥 − 7)(10𝑥 + 29𝑥 − 21)is a perfect square by showing that it is the square of a polynomial. 8𝑛+1 2 4𝑛+1 2𝑛+2 4𝑛+2 8) Factor 90𝑎 𝑏 − 25𝑎 𝑏 − 240𝑎𝑏 , where 𝑛is a positive integer. Part 2 Choose at least 2 problems below to set up and solve. 10) If a number is added to its square, the result is 56. Find the number. 11) A positive number is 30 less than its square. Find the number. 12) Find two consecutive odd integers whose product is 143. 13) The length of a rectangle is 8 cm greater than its width. Find the dimensions of the 2 rectangle if its area is 105 𝑐𝑚. 2 14) Find the dimensions of a rectangle whose perimeter is 46 m and whose area is126 𝑚. (let 𝑤 be the width) 15) The sum of two numbers is 25 and the sum of their squares is 313. Find the numbers. 16) Vanessa built a rectangular pen for her dogs. She used an outside wall of the garage for one of the sides of the pen. She had to buy 20 m of fencing in order to build the other sides of the pen. Find the dimensions of the pen if the area is 2 48 𝑚. Part 2 Choose at least 2 problems below to set up and solve 9) A garden plot 4 m by 12 m has one side along a fence as shown in the diagram to the right. The area of the garden is to be doubled by digging a border of uniform width on the other three sides. What should the width of the border be? 10) A box has a square bottom and top and is 5 cm high. Find its volume if its total surface area 2 is 192 𝑐𝑚. Optional Challenge: 11) The width of a box is 4 inches more than the height. The length is the difference of 9 inches and the height. a. Write a polynomial that represents the volume of the box. b. The volume of the box is 180 cubic inches. What are all the possible dimensions of the box? c. Which dimensions result in a box with the smallest possible surface area? Explain your reasoning. 𝑣 1) 𝐶 = 2π𝑟;𝑟 2) 𝑠 = 𝑟 ;𝑟 3) 𝑙 = 𝑃𝑟𝑡;𝑃 2 1 4) 𝐴 = 2𝑎 + 4𝑎ℎ;ℎ 5) 𝐴 = 2 ℎ(𝑎 + 𝑏);ℎ 6) 𝑝 = 2(𝑙 + 𝑤);𝑤 𝑥+𝑦 𝑛 𝑣−𝑢 7) 𝑚 = 2 ;𝑦 8) 𝑆 = 2 (𝑎 + 𝑙);𝑎 9) 𝑎 = 𝑡 ;𝑡 2 2 𝑎 𝑓𝑔 10) 𝑣 =𝑢 + 2𝑎𝑠;𝑠 11) 𝑆 = 𝑎−𝑟 ;𝑟 12) 𝐹 = 𝑓+𝑔−𝑑 ;𝑑 Challenge: (these may require factoring) 13) 𝑎 = 180(𝑛−2) 𝑛 ;𝑛 14) 𝑟 = 𝑎𝑏 𝑎+𝑏 ;𝑎 15) 𝐶 = 𝐾 ( );𝑅 𝑅𝑟 𝑅−𝑟 ________________________________________________________ Additional Practice: (optional) 𝑚 1) 𝐹 = 𝑚𝑎;𝑎 2) 𝑑 = 𝑣 ;𝑣 3) 𝐴 = 𝑃 + 𝑃𝑟𝑡;𝑡 2 𝑛 4) 𝑠 = 𝑣𝑡 + 16𝑡 ;𝑣 5) 𝑆 = 2 (𝑎 + 𝑙);𝑛 6) 𝐴 = 𝑃(𝑙 + 𝑟𝑡);𝑟 𝑣−𝑢 5 𝑥+𝑦+𝑧 7) 𝑎 = 𝑡 ;𝑣 8) 𝐶 = 9 (𝐹 − 32);𝐹 9) 𝑚 = 3 ;𝑦  𝑛 𝑎−𝑟𝑙 10) 𝑠 = 2 (𝑎 + 𝑙);𝑙 11) 𝑙 = 𝑎 + (𝑛 − 1)𝑑;𝑛 12) 𝑆 = 1−𝑟 ;𝑙 Challenge: (these may require factoring) 𝑟 𝑓𝑔 3𝑛+2 13) 𝑆 = 1−𝑟 ;𝑟 14) 𝐹 = 𝑓+𝑔−𝑑 ;𝑓 15) 𝐾 = 𝑛+1 ;𝑛 2 3 3 1) 𝑥 + 5 = 2 2) (𝑥 − 1) = 36 3) 𝑥 − 4 = 2 4) (𝑥 + 1) =− 8 2 3 1 3 3 5) 2(𝑥 − 4) = 50 6) (𝑥 − 1) = 8 7) 3(5𝑥 − 2) = 81 8) 5(𝑥 − 1) = 10 3 2 2 −3 2 2 −3 9) − (2𝑥 + 1) = 64 10) 3 (𝑥 + 5) = 6 11) 𝑥 = 27 12) 𝑥 = (3) Solving absolute value 13) |𝑥| = 15 14) |𝑥| =− 19 15) |𝑥 − 7| = 11 |2𝑥−6| 𝑥 16) 3 = 10 17) 3 | 9 |+7=8 18) 6 − 3 | − 8𝑟 − 9| = − 15 Challenge: do in notebook for more space… Solve for the missing variable 2 2 −3 3 19) (𝑟 − 1) =1 20) (2𝑥 − 6) + 9 = 13 4 1 1 −3 2 4 2 21) − 8 + (𝑥 − 9) = 248 22) 4(𝑦 − 𝑦 + 1) = (16𝑦) 23) 6|2𝑥 + 3| − 7 = 2|2𝑥 + 3| + 1 24) ||𝑥 − 1| − 6| − 1 = 2 3 3 2 25) 𝑥 − 3𝑥 + 2 = 𝑥 − 1