6-4 Factoring and Solving Polynomial Equations (Day2)

Download Report

Transcript 6-4 Factoring and Solving Polynomial Equations (Day2)

6 – 4 Factoring and Solving Polynomial Equations Day 2

Objective: Use factoring to solve polynomial equations.

Example 5 Solve 2 2

x

5

x

5  24

x

 14

x

3  24

x

 14

x

3 Given 2 2 2 2

x

5  4 2 2 14   

x

3 7

x

3 3    2 

x

 2 24 12 

x

4    0 0   0

x

 2 

x

 2   0

2 2  3  

x

 2 

x

 2   0 2

x

 0 or

x

2 0 or

x

0 or

x x

 0,

x

2  3

x

  2

x

 2

x

  3 0

Example 5 An optical company is going to make a glass prism that has a volume of 15 cm 3 . The height of the prism is h cm, and the base will be a right triangle with legs of length (h – 2) and (h – 3) cm. What will be the height of the prism?

Solution If a prism has a volume of V cubic units, a base with an area of B square units, and a height of h units, then V = Bh Base of a triangular prism is B = ½ leg 1 x leg 2

V

Bh

15

cm

3     1 2 

h

 2 

h

 3    

h

    1 2 

h

2  5

h

 6    

h

30

cm

  1 2 

h

3

h

3  5

h

2  5

h

2  6

h

 6

h

 0 0 0   

h

3  5

h

2

h

2 

h

2 

h

 6    6

h h

 5   30

h

 5 

h

2

0 or

h h

2  

6

h

5

h

  

6

h

 

i

6 0

Since distance cannot be an imaginary number the height must be h = 5 cm.

Homework page 348 28 –60 even, 75 – 85 odd