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Applying Factoring Chapter 10. Solve.  (x – 3)(x – 4) = 0.

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Presentation on theme: "Applying Factoring Chapter 10. Solve.  (x – 3)(x – 4) = 0."— Presentation transcript:

1 Applying Factoring Chapter 10

2 Solve.  (x – 3)(x – 4) = 0.

3 Solve.  (x – 3)(x – 4) = 0.

4 Solve  x 2 + 5x + 6 = 0

5 Solve  x 2 + 5x + 6 = 0

6 Solve.  x 2 – 3 = 2x

7 Solve.  x 2 – 3 = 2x

8 Solve.  (x + 2)(x + 3) = 12

9 Solve.  (x + 2)(x + 3) = 12

10 Simplify.  (2x 2 - 4) - (x 2 + 3x - 3)

11 Simplify.  (2x 2 - 4) - (x 2 + 3x - 3)

12 Factor.  9y 2 - 49

13 Factor.  9y 2 - 49

14 Solve.  x 2 – 5x = 0

15 Solve.  x 2 – 5x = 0

16 Simplify.  (3x 2 - 4x + 6) - (-2x 2 - 3x - 9)

17 Simplify.  (3x 2 - 4x + 6) - (-2x 2 - 3x - 9)

18 Solve.  x 2 – 4 = 0

19 Solve.  x 2 – 4 = 0

20 Simplify.  (4x 2 – 4x – 7)(x + 3)

21 Simplify.  (4x 2 – 4x – 7)(x + 3)

22 Factor  3x 2 - 5x - 2

23 Factor  3x 2 - 5x - 2

24 Solve.  (x – 5) 2 – 100 = 0

25 Solve.  (x – 5) 2 – 100 = 0

26 Simplify.  –3x(4x 2 – x + 10)

27 Simplify.  –3x(4x 2 – x + 10)

28 Factor.  5m 2 + 13m - 6

29 Factor.  5m 2 + 13m - 6

30 Solve.  The room that is shown in the figure below has a floor space of 2x² + x - 15 square feet. If the width of the room is (x + 3) feet, what is the length? x + 3

31 Solve.  The room that is shown in the figure below has a floor space of 2x² + x - 15 square feet. If the width of the room is (x + 3) feet, what is the length? x + 3

32 Solve.  x 2 +3x = 0

33 Solve.  x 2 +3x = 0

34 Simplify.  (3x 3 + 3x 2 – 4x + 5) + (x 3 – 2x 2 + x – 4)

35 Simplify.  (3x 3 + 3x 2 – 4x + 5) + (x 3 – 2x 2 + x – 4)

36 Solve.  2x 2 – 6 = x

37 Solve.  2x 2 – 6 = x

38 Solve.  2x(x+1) = 7x – 2

39 Solve.  2x(x+1) = 7x – 2

40  Homework Homework


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