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Factoring Differences of Squares - Lesson Summary and Homework

This lesson focuses on factoring binomials that are the difference of squares. It covers examples, vocabulary, and practice problems. Homework includes various exercises to reinforce the concept.

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Factoring Differences of Squares - Lesson Summary and Homework

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  1. Transparency 3-1 5-Minute Check on Chapter 2 • Evaluate 42 - |x - 7| if x = -3 • Find 4.1  (-0.5) • Simplify each expression • 3. 8(-2c + 5) + 9c 4. (36d – 18) / (-9) • A bag of lollipops has 10 red, 15 green, and 15 yellow lollipops. If one is chosen at random, what is the probability that it is notgreen? • Which of the following is a true statement Standardized Test Practice: 8/4 < 4/8 -4/8 < -8/4 -4/8 > -8/4 -4/8 > 4/8 A B C D Click the mouse button or press the Space Bar to display the answers.

  2. Lesson 9-5 Factoring Differences of Squares

  3. Transparency 5 Click the mouse button or press the Space Bar to display the answers.

  4. Transparency 5a

  5. Objectives • Factor binomials that are the difference of squares • Solve equations involving the difference of squares

  6. Vocabulary • NA –

  7. Difference of Squares • Symbols: a2 – b2 = (a + b) (a – b) or (a – b) (a + b) • Examples: • x2 – 9 = (x – 3) (x + 3) or (x + 3) (x – 3) • b2 – 25 = (b – 5) (b + 5) or (b + 5) (b – 5)

  8. A. Factor . B. Factor . Write in form and Factor the difference of squares. Factor the difference of squares. Answer: Answer: Example 1

  9. Factor The GCF of and 27b is 3b. and Factor the difference of squares. Answer: Example 2

  10. Factor The GCF ofand 2500 is 4. and Factor the difference of squares. and Factor the difference of squares. Answer: Example 3

  11. Factor Original Polynomial Factor out the GCF. Group terms with common factors. Factor each grouping. is the common factor. Factor the difference of squares,into . Answer: Example 4

  12. A. Solve by factoring. Answer: The solution set is Original equation. and Factor the difference of squares. Zero Product Property or Solve each equation. Example 5a

  13. Answer: The solution set is Solve by factoring. Original equation Subtract 3y from each side. The GCF of and 3y is 3y. and or or Example 5b Applying the Zero Product Property, set each factor equal to zero and solve the resulting three equations.

  14. Summary & Homework • Summary: • Difference of Squares: a2 – b2 = (a + b)(a - b) or (a – b)(a + b) • Sometimes it may be necessary to use more than one factoring technique or to apply a factoring more than once • Homework: • Pg. 505 16-30 even, 34,36

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