This document discusses solving quadratic equations by factoring. It begins by reviewing quadratic equations and factoring polynomials. Examples are then provided of solving quadratic equations in standard form by factoring the left-hand side of the equation, setting each factor equal to zero using the zero product property, and solving the resulting linear equations. Three word problems involving projectile motion are worked through using this process. The steps for solving quadratic equations by factoring are recapped at the end.

1. QUADRATIC
EQUATIONS

2. • Review Quadratic Equations
• Recall the different techniques of
factoring polynomials
• Solve quadratic Equations by factoring
By the end of this lesson...
• Show focus, collaboration, and problem-
solving skill

4. ²
≠

5. ²
≠
quadratic

6. ²

7. ² 2

8. ²

9. ²
standard

10. ²

11. ²
positive

13. roots

15. two

17. parabola

18. FACTORING
POLYNOMIALS

19. Fill in the blanks with the correct factors
of each polynomial.
1. 2x² + 8x = 2x(_____)
2. s² + 8s + 12 = (s + 2)(____)
3. x² – 10x + 25 = (____)(x – 5)
4. 9t² – 4 = (______)(3t – 2)
5. 2x² + 3x – 14 = (2x + 7)(____)

20. Fill in the blanks with the correct factors
of each polynomial.
1. 2x² + 8x = 2x(_____)
2. s² + 8s + 12 = (s + 2)(____)
3. x² – 10x + 25 = (____)(x – 5)
4. 9t² – 4 = (______)(3t – 2)
5. 2x² + 3x – 14 = (2x + 7)(____)
x+4
s+6
x-5
3t+2
x-2

21. Fill in the blanks with the correct factors
of each polynomial.
1. 2x² + 8x = 2x(_____)
2. s² + 8s + 12 = (s + 2)(____)
3. x² – 10x + 25 = (____)(x – 5)
4. 9t² – 4 = (______)(3t – 2)
5. 2x² + 3x – 14 = (2x + 7)(____)
x+4
s+6
x-5
3t+2
x-2
≠

22. QUADRATIC
EQUATIONS

23. Some quadratic equation in standard form can be
transformed in to factored form. Such equations
can be solved using the “Zero Product Property”

24. Some quadratic equation in standard form can be
transformed in to factored form. Such equations
can be solved using the “Zero Product Property”
Standard Form: x² + 3x - 10 = 0
Factored Form: (x + 5)(x – 2) =
0
Therefore: (x + 5) = 0 or (x – 2) =
0

25. Some quadratic equation in standard form can be
transformed in to factored form. Such equations
can be solved using the “Zero Product Property”
Standard Form: x² + 3x - 10 = 0
Factored Form: (x + 5)(x – 2) =
0
Therefore: (x + 5) = 0 or (x – 2) = 0
x = -
5
x =
2

26. STEPS:
1. Write the equation in standard form.
2. Factor the left-hand side of the equation.
3. Set each factor equal to zero using
the Zero Product Property.
4. Solve each linear equation.

27. EXAMPLES

28. Solve: 3x² = - 18x

29. Solve: 3x² = - 18x

30. Solve: 11x² - 13x = 8x – 3x²

31. Solve: 11x² - 13x = 8x – 3x²

32. Solve: 2x² – 5x = 3

33. Solve: 2x² – 5x = 3
The roots are -1/2 and 3

34. Solve: 2x² – 5x = 3

35. Solve: (x + 1)(8x + 1) = 18x
The roots are 1/8 and 1

41. A relief package is released from a helicopter at 1600 feet. The height
of the package can be modeled by the equation: h = -16t² + 1600, where
h is the height of the package in feet and t is the time in seconds.
The pilot wants to know how long it will take for the package to hit the
ground.

42. A relief package is released from a
helicopter at 1600 feet. The height
of the package can be modeled by the
equation: h = -16t + 1600, where h is
the height of the package in feet and
t is the time in seconds.
The pilot wants to know how long it
will take for the package to hit the
ground.
2
h = -16t + 1600
2
0 = -16t + 1600
2
16t - 1600 = 0
2
16(t - 100) = 0
2
16(t + 10)(t - 10) = 0
t + 10 = 0 t - 10 = 0
t = -10 t = 10
Answer: 10 seconds

43. Robert threw a rock off a bridge into the river. The distance from the
rock to the river is modeled by the equation h = -16t² - 16t + 60, where
h is the height in ft. and t is the time in seconds. How long will it take
the rock to hit the surface?

44. Robert threw a rock off a bridge
into the river. The distance from
the rock to the river is molded by
the equation h = -16t - 16t + 60,
where h is the height in ft. and t is
the time in seconds.
How long will it take the rock to hit
the surface?
2
h = -16t - 16t + 60
2
0 = -16t - 16t + 60
2
16t + 16 - 60 = 0
2
4(4t + 4 - 15) = 0
2
4(2t - 3)(2t + 5) = 0
2t - 3 = 0 2t + 5 = 0
2t = 3 2t = -5
Answer: 3/2 or 2.5 seconds
t = 3/2 t = -5/2

45. During a game of golf, Kyle hits his
ball out of a sand trap. The height of
the golf ball is modeled by the
equation h = -16t² + 20t - 4, where h
is the height in feet and t is the time
in seconds since the ball was hit.
How long does it take for the golf ball
to hit the ground?

46. During a game of golf, Kyle hits his
ball out of a sand trap. The height of
the golf ball is modeled by the
equation h = -16t² + 20t - 4, where h
is the height in feet and t is the time
in seconds since the ball was hit.
How long does it take for the golf ball
to hit the ground?
h = -16t + 20t - 4
2
0 = -16t + 20t - 4
2
16t - 20t + 4 = 0
2
4(4t - 5t + 1) = 0
2
4(4t - 1)(t - 1) = 0
4t - 1 = 0 t - 1 = 0
4t = 1 t = 1
Answer: 1 second
t = 1/4

47. RECAP

48. 1. Write the equation in ___________.
2. ______ the left-hand side of the equation.
3. Set each factor ___________ using the Zero
Product Property.
4. Solve each linear equation.
RECALL: Solving Quadratic Equations by Factoring

49. 1. Write the equation in standard form.
2. Factor the left-hand side of the equation.
3. Set each factor equal to zero using the Zero
Product Property.
4. Solve each linear equation.
RECALL: Solving Quadratic Equations by Factoring