Unit 10 Quadratic and Factoring by b7O6Dpvr

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									Unit 10 Quiz 2A                                                     Name:_____________________

              Factor the expression.

1. (A1.5.C)    d2 + 10d + 9




2. (A1.5.C) k2 + kf – 2f2




3. (A1.5.C)




4. (A1.5.C) 4x2 – 81y2




5. (A1.5.C)




6. (A1.5.C) 16m2 – 24mn + 9n2




              Solve the equation using the zero-product property.

7. (A1.5.C)




8. (A1.5.C)




9. (A1.5.C)
10.(A1.5.C)




11.(A1.5.C)




12.(A1.5.C) Tasha is planning an expansion of a square flower garden in a city park. If each side of the original garden is
            increased by 7 m, the new total area of the garden will be 144 m2. Find the length of each side of the original
            garden.




13.(A1.5.C) The area of a playground is 336 yd2. The width of the playground is 5 yd longer than its length. Find the
            length and width of the playground.




14.(A1.5.C)Explain how to factor the following trinomial.
              g2 + 4g – 60
Unit 10 Quiz 2

Answer Section

SHORT ANSWER

       1. ANS:
          (d + 9)(d + 1)

           PTS:   1              DIF: L3            REF: 9-5 Factoring Trinomials of the Type x^2 + bx + c
           OBJ:   9-5.1 Factoring Trinomials        NAT: NAEP 2005 A3c | ADP J.1.4
           STA:   WA 1.5.5       TOP: 9-5 Example 1                     KEY: polynomial | factoring trinomials

       2. ANS:
          (k + 2f)(k – f)

           PTS:   1              DIF: L3            REF: 9-5 Factoring Trinomials of the Type x^2 + bx + c
           OBJ:   9-5.1 Factoring Trinomials        NAT: NAEP 2005 A3c | ADP J.1.4
           STA:   WA 1.5.5       TOP: 9-5 Example 3                     KEY: polynomial | factoring trinomials

       3. ANS:


           PTS:   1              DIF: L2               REF: 9-7 Factoring Special Cases
           OBJ:   9-7.1 Factoring Perfect-Square Trinomials                  NAT: ADP J.1.4
           STA:   WA 1.5.5       TOP: 9-7 Example 1
           KEY:   polynomial | factoring trinomials | perfect-square trinomial

       4. ANS:
          (2x + 9y)(2x – 9y)

           PTS:   1              DIF: L3               REF: 9-7 Factoring Special Cases
           OBJ:   9-7.2 Factoring the Difference of Squares                 NAT: ADP J.1.4
           STA:   WA 1.5.5       TOP: 9-7 Example 4
           KEY:   polynomial | factoring trinomials | difference of squares

       5. ANS:


           PTS:   1              DIF: L3               REF: 9-7 Factoring Special Cases
           OBJ:   9-7.1 Factoring Perfect-Square Trinomials                  NAT: ADP J.1.4
           STA:   WA 1.5.5       TOP: 9-7 Example 2
           KEY:   polynomial | factoring trinomials | perfect-square trinomial

       6. ANS:
     (4m – 3n)2

     PTS:   1              DIF: L3               REF: 9-7 Factoring Special Cases
     OBJ:   9-7.1 Factoring Perfect-Square Trinomials                  NAT: ADP J.1.4
     STA:   WA 1.5.5       TOP: 9-7 Example 2
     KEY:   polynomial | factoring trinomials | perfect-square trinomial

 7. ANS:
    x = –1 or x = 1

     PTS:   1              DIF: L2              REF: 10-4 Factoring to Solve Quadratic Equations
     OBJ:   10-4.1 Solving Quadratic Equations
     NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
     STA:   WA 1.5.6 | WA 2.2.2bTOP:            10-4 Example 1
     KEY:   zero-product property | solving quadratic equations

 8. ANS:
                     1
     n = 0 or n =
                     10

     PTS:   1              DIF: L2              REF: 10-4 Factoring to Solve Quadratic Equations
     OBJ:   10-4.1 Solving Quadratic Equations
     NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
     STA:   WA 1.5.6 | WA 2.2.2bTOP:            10-4 Example 1
     KEY:   zero-product property | solving quadratic equations

 9. ANS:
    z = –3 or z = 9

     PTS:   1               DIF: L2              REF: 10-4 Factoring to Solve Quadratic Equations
     OBJ:   10-4.1 Solving Quadratic Equations
     NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
     STA:   WA 1.5.6 | WA 2.2.2bTOP:             10-4 Example 2
     KEY:   factoring | solving quadratic equations

10. ANS:
    z = 1 or z = –2

     PTS:   1               DIF: L2              REF: 10-4 Factoring to Solve Quadratic Equations
     OBJ:   10-4.1 Solving Quadratic Equations
     NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
     STA:   WA 1.5.6 | WA 2.2.2bTOP:             10-4 Example 2
     KEY:   factoring | solving quadratic equations

11. ANS:
    c = 0 or c = 4

     PTS:   1             DIF:   L2            REF:   10-4 Factoring to Solve Quadratic Equations
             OBJ:   10-4.1 Solving Quadratic Equations
             NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
             STA:   WA 1.5.6 | WA 2.2.2bTOP:             10-4 Example 2
             KEY:   factoring | solving quadratic equations

        12. ANS:
            5m

             PTS:   1               DIF: L2              REF: 10-4 Factoring to Solve Quadratic Equations
             OBJ:   10-4.1 Solving Quadratic Equations
             NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
             STA:   WA 1.5.6 | WA 2.2.2bTOP:             10-4 Example 4
             KEY:   factoring | solving quadratic equations | word problem | problem solving

        13. ANS:
            length = 16 yd, width = 21 yd

             PTS:   1               DIF: L3              REF: 10-4 Factoring to Solve Quadratic Equations
             OBJ:   10-4.1 Solving Quadratic Equations
             NAT:   NAEP 2005 A4a | NAEP 2005 A4c | ADP J.3.5 | ADP J.5.3
             STA:   WA 1.5.6 | WA 2.2.2bTOP:             10-4 Example 4
             KEY:   factoring | solving quadratic equations | word problem | problem solving

ESSAY

        14. ANS:
            [4]  Since the second sign is negative, one of the factors of 60 will be positive and one will
                 be negative. Find two factors of 60 that have a difference of 4. 10 – 6 = 4; Since the
                 first sign is positive, 10 is positive and 6 is negative. The factors of g2 will be g.
                 (g+ 10)(g – 6)
            [3]  correct explanation with one minor factoring error
            [2]  correct explanation with one error in the signs of the factors
            [1]  correct factors with no explanation


             PTS:   1              DIF: L3             REF: 9-5 Factoring Trinomials of the Type x^2 + bx + c
             OBJ:   9-5.1 Factoring Trinomials         NAT: NAEP 2005 A3c | ADP J.1.4
             STA:   WA 1.5.5       TOP: 9-5 Example 3
             KEY:   extended response | rubric-based question | polynomial | factoring trinomials

								
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