Beginning Algebra, 11th Edition

412 CHAPTER 6 Factoring and Applications

6.4 Special Factoring Techniques

Difference of Squares

Perfect Square Trinomials

Difference of Cubes

Sum of Cubes

x^3 y^3  1 xy 21 x^2 xyy^22

x^3 y^3  1 xy 21 x^2 xyy^22

x^2  2 xyy^2  1 xy 22

x^2  2 xyy^2  1 xy 22

x^2 y^2  1 xy 21 xy 2

Factor.

= 1 m- 221 m^2 + 2 m+ 42 = 1 z+ 321 z^2 - 3 z+ 92

=m^3 - 23 =z^3 + 33

m^3 - 8 z^3 + 27

= 13 x+ 122 = 12 x- 522

9 x^2 + 6 x+ 1 4 x^2 - 20 x+ 25

= 12 x+ 3212 x- 32

4 x^2 - 9

CONCEPTS EXAMPLES

6.5 Solving Quadratic Equations

by Factoring

Zero-Factor Property

If aand bare real numbers and if then or

Solving a Quadratic Equation by Factoring

Step 1 Write the equation in standard form.

Step 2 Factor.

Step 3 Use the zero-factor property.

Step 4 Solve the resulting equations.

Step 5 Check.

b=0.

ab=0, a= 0 If then or

Solve

or

Both solutions satisfy the original equation. The solution set is

E-^32 , 5F.

x=-

3

2

2 x=- 3 x= 5

2 x+ 3 = 0 x- 5 = 0

12 x+ 321 x- 52 = 0

2 x^2 - 7 x- 15 = 0

2 x^2 = 7 x+15.

1 x- 221 x+ 32 =0, x- 2 = 0 x+ 3 =0.

6.6 Applications of Quadratic Equations

Pythagorean Theorem

In a right triangle, the square of the hypotenuse equals the
sum of the squares of the legs.

a^2 b^2 c^2

In a right triangle, one leg measures 2 ft longer than the other. The

hypotenuse measures 4 ft longer than the shorter leg. Find the lengths

of the three sides of the triangle.

Let the length of the shorter leg. Then

Verify that the solutions of this equation are and 6. Discard as

a solution. Check that the sides have lengths

6 ft, 6 + 2 = 8 ft, and 6 + 4 = 10 ft.

- 2 - 2

x^2 + 1 x+ 222 = 1 x+ 422.

x=

Hypotenuse

Leg a c

Leg b

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