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					          Factoring to Solve Quadratic
ALGEBRA 1 LESSON 10-5

                                                        (For help, go to Lessons 2-2 and 9-6.)


          Solve and check eachEquations
                              equation.
          1. 6 + 4n = 2             2. a – 9 = 4                 3. 7q + 16 = –3
                                       8


          Factor each expression.
          4. 2c2 + 29c + 14         5. 3p2 + 32p + 20            6. 4x2 – 21x – 18




                                           5-13
       Factoring to Solve Quadratic
ALGEBRA 1 LESSON 10-5




Solutions       Equations
          1. 6 + 4n = 2
                 4n = –4
                  n = –1
          Check: 6 + 4(–1) = 6 + (–4) = 2
              a
          2. 8 – 9 = 4
                a
                8
                   = 13
                   a = 104
          Check: 104 – 9 = 13 – 9 = 4
                        8
          3. 7q + 16 = –3
                  7q = –19
                   q = –2 5
                           7
          Check: 7 (–2 5 ) + 16 = 7(– 19 ) + 16 = –19 + 16 = –3
                       7               7


                                            5-13
       Factoring to Solve Quadratic
ALGEBRA 1 LESSON 10-5




Solutions (continued) Equations
          4. 2c2 + 29c + 14 = (2c + 1)(c + 14)
             Check: (2c + 1)(c + 14) = 2c2 + 28c + c + 14 = 2c2 + 29c + 14

          5. 3p2 + 32p + 20 = (3p + 2)(p + 10)
             Check: (3p + 2)(p + 10) = 3p2 + 30p + 2p + 20 = 3p2 + 32p + 20

          6. 4x2 – 21x – 18 = (4x + 3)(x – 6)
             Check: (4x + 3)(x – 6) = 4x2 – 24x + 3x – 18 = 4x2 – 21x – 18




                                         5-13
          Factoring to Solve Quadratic
ALGEBRA 1 LESSON 10-5




                        Equationsan open-top box. The total
               The diagram shows a pattern for
               area of the sheet of materials used to make the box is 130 in.2. The
               height of the box is 1 in. Therefore, 1 in.  1 in. squares are cut from
               each corner. Find the dimensions of the box.


                            Define: Let x = width of a side of the box.
                                    Then the width of the material = x + 1 + 1 = x + 2
                                    The length of the material = x + 3 + 1 + 1 = x + 5



                            Relate: length  width = area of the sheet




                                            5-13
          Factoring to Solve Quadratic
ALGEBRA 1 LESSON 10-5




                   Equations(continued)

         Write: (x + 2) (x + 5) = 130

                        x2 + 7x + 10 = 130         Find the product (x + 2) (x + 5).

                   x2 + 7x – 120 = 0               Subtract 130 from each side.

                 (x – 8) (x + 15) = 0              Factor x2 + 7x – 120.

           x–8=0             or     x + 15 = 0     Use the Zero-Product Property.

          x=8                or     x = –15        Solve for x.

          The only reasonable solution is 8. So the dimensions of the box are
          8 in.  11 in.  1 in.


                                                 5-13
          Factoring to Solve Quadratic
ALGEBRA 1 LESSON 10-5




                                Equations
          1. Solve (2x – 3)(x + 2) = 0.
                        3
                 –2, 2
          Solve by factoring.
          2. 6 = a2 – 5a        3. 12x + 4 = –9x2   4. 4y2 = 25
                                         2                   5
                   –1, 6             –                   ±
                                         3                   2




                                    5-13

				
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