Solve polynomials by factoring

1
Section 7.6

• Solve polynomials by factoring

2
Quadratic Equation

• A quadratic equation in x is an equation that can
  be written in the standard form
• ax2 bx c 0

3
Quadratic Equation

• A quadratic equation in x is an equation that can
  be written in the standard form
• ax2 bx c 0
• Example 3x2 7x 4 0

4
Zero-Product Principle

• If the product of 2 algebraic expressions is
  zero, then at least one of those factors equals
  zero.
• If AB 0, then A 0 or B 0

5
Zero-Product Principle

• For example, consider the equation
• (x 4)(x 3) 0

6
Zero-Product Principle

• For example, consider the equation
• (x 4)(x 3) 0
• This means
• x 4 0 or x 3 0

7
Zero-Product Principle

• For example, consider the equation
• (x 4)(x 3) 0
• This means
• x 4 0 or x 3
  0
• x 4

8
Zero-Product Principle

• For example, consider the equation
• (x 4)(x 3) 0
• This means
• x 4 0 or x 3
  0
• x 4 or x 3

9
Solving by Factoring

• Rewrite the equation with all terms on one side.
• Factor completely.
• Apply the zero-product principle, setting each
  factor equal to zero.
• Solve the equations from step 3.
• Check your work.

10

• Solve x2 2x 15 0

11

• Solve x2 2x 15 0
• x2 2x 15 0 Factor
• c is negative, b is negative, 15 53 and 35-2
• (x 3)(x 5) 0 Use zero-product
  principal
• x 3 0 or
  x 5 0
• x 3 or
  x 5
• So the solution set is 3,5

12

• Solve 2x2 5x 3 0

13

• Solve 2x2 5x 3 0
• 2x2 5x 3 0 Factor
• c is negative, b is negative
• (2x )(x ) 0 Fill in the rest with trial
  and error
• (2x 1)(x 3) 0 Use zero-product
  principal
• 2x 1 0 or
  x 3 0
• 2x 1 or
  x 3
• x (-1/2)
• So the solution set is 1/2,3

14

• Solve 2x2 5x

15

• Solve 2x2 5x
• 2x2 5x Get all terms on one side
• 2x2 5x 0 Factor out GCD
• x(2x - 5) 0 zero-product principal
• x 0 or 2x
  5 0
• 2x 5
• x (5/2)
• So the solution set is 0,5/2

16

• Solve 9x2 100

17

• Solve 9x2 100
• 9x2 100 Get all terms on one side
• 9x2 100 0 Factor out GCD
• (3x 10)(3x 10)0 Diff of squares A 3x B10
• apply zero-product principal
• 3x 10 0 or 3x 10
  0
• 3x -10 3x 10
• x -(10/3) x (10/3)
• So the solution set is -10/3,10/3

18

• A rectangular parking lot has a length that is 3
  yards greater than the width. The area of the
  parking lot is 180 square yards. Find the length
  and width.

19

• A rectangular parking lot has a length that is 3
  yards greater than the width. The area of the
  parking lot is 180 square yards. Find the length
  and width.
• x width x 3 length
• width length Area
• (x)(x 3) 180
• x2 3x 180 0
• (x 15)(x 12) 0
• x 15 0 or x 12
  0
• x 15 x 12
  
• x -15 makes no sense width 12 yards
• length 15 yards

20

• A vacant rectangular lot is being turned into a
  community garden measuring 15 meters by 12
  meters. A path of uniform width is to surround
  the garden. If the total area is 378 square
  meters, find the width of the path.

21

• A vacant rectangular lot is being turned into a
  community garden measuring 15 meters by 12
  meters. A path of uniform width is to surround
  the garden. If the total area is 378 square
  meters, find the width of the path.
• x width of path
• 4x2 54x 180 378
• 4x2 54x 198 0
• 2(2x2 27x 99) 0
• 2(2x 33)(x 3) 0
• 2x 33 0 or x 3
  0
• x -33/2 makes no sense x 3