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# Algebra I Chapter 8

Algebra I Chapter 8

## CHAPTER8Study Guideand

CHAPTER8Study Guideand ReviewDownload VocabularyReview from ca.algebra1.comBe sure the followingKey Concepts are notedin your Foldable. Key ConceptsMonomials and Factoring (Lesson 8-1)• The greatest common factor (GCF) of two ormore monomials is the product of their commonprime factors.Factoring Using the Distributive Property(Lesson 8-2)• Using the Distributive Property to factorpolynomials with four or more terms is calledfactoring by grouping.ax + bx + ay + by = x(a + b) + y(a + b)= (a + b)(x + y)• Factoring can be used to solve some equations.According to the Zero Product Property, for anyreal numbers a and b, if ab = 0, then eithera = 0, b = 0, or both a and b equal zero.Factoring Trinomials and Differencesof Squares (Lessons 8-3, 8-4, and 8-5)• To factor x 2 + bx + c, find m and n with asum of b and a product of c. Then writex 2 + bx + c as (x + m)(x + n).• To factor a x 2 + bx + c, find m and n with aproduct of ac and a sum of b. Then write asax 2 + mx + nx + c and factor by grouping.a 2 - b 2 = (a + b)(a - b) or (a - b)(a + b)Perfect Squares and Factoring (Lesson 8-6)• a 2 + 2ab + b 2 = (a + b )2anda 2 - 2ab + b 2 = (a - b )2• For a trinomial to be a perfect square, the first andlast terms must be perfect squares, and the middleterm must be twice the product of the squareroots of the first and last terms.• For any number n > 0, if x 2 = n, then x = ± √ n .Key Vocabularycomposite number (p. 420)factored form (p. 421)factoring (p. 426)factoring by grouping(p. 427)greatest common factor(p. 422)perfect square trinomials(p. 454)prime factorization (p. 421)prime number (p. 420)prime polynomial (p. 443)roots (p. 428)Vocabulary CheckState whether each sentence is true or false.If false, replace the underlined word,phrase, expression, or number to makea true sentence.1. The number 27 is an example of a primenumber.2. 2x is the greatest common factor of 12 x 2and 14xy.3. 66 is an example of a perfect square.4. 61 is a factor of 183.5. The prime factorization of 48 is 3 · 4 2 .6. x 2 - 25 is an example of a perfect squaretrinomial.7. The number 35 is an example of acomposite number.8. x 2 - 3x - 70 is an example of a primepolynomial.9. Expressions with four or more unliketerms can sometimes be factored bygrouping.10. (b - 7)(b + 7) is the factorization of adifference of squares.Vocabulary Review at ca.algebra1.comChapter 8 Study Guide and Review 461

CHAPTER8Study Guide and ReviewLesson-by-Lesson Review8–1 Monomials and Factoring (pp. 420–424)Factor each monomial completely.11. 28n 3 12. -33a 2 b13. 150st 14. -378pq 2 r 2Find the GCF of each set of monomials.15. 35, 30 16. 12, 18, 4017. 12ab, 4a 2 b 2 18. 16mrt, 30m 2 r19. 20n 2 , 25np 520. 60x 2 y 2 , 15xyz, 35xz 321. HOME IMPROVEMENT A landscapearchitect is designing a stone path tocover an area 36 inches by 120 inches.What is the maximum size square stonethat can be used so that none of thestones have to be cut?Example 1 Factor 68cd 2 completely.68cd 2 = 4 · 17 · c · d · d= 2 · 2 · 17 · c · d · d 4 = 2 · 2Thus, 68 cd 2 in factored form is2 · 2 · 17 · c · d · d.68 = 4 · 17, d 2 = d · dExample 2 Find the GCF of 15x 2 y and45xy 2 .15x 2 y = 3 · 5 · x · x · yFactor each number.45xy 2 = 3 · 3 · 5 · x · y · y Circle the commonprime factors.The GCF is 3 · 5 · x · y or 15xy.8–2Factoring Using the Distributive Property (pp. 426–431)Factor each polynomial.22. 13x + 26y 23. a 2 - 4ac + ab -4bc24. 24a 2 b 2 - 18ab 25. 26ab + 18ac + 32 a 226. 4rs + 12ps + 2mr + 6mp27. 24am - 9an + 40bm - 15bnSolve each equation. Check the solutions.28. x(2x - 5) = 029. 4x 2 = -7x30. (3n + 8)(2n - 6) = 031. EXERCISE A gymnast jumps on atrampoline traveling at 12 feet persecond. Her height h in feet above thetrampoline after t seconds is given bythe formula h = 12t - 16t 2 . How long isthe gymnast in the air before returningto the trampoline?Example 3 Factor 2x 2 - 3xz - 2xy + 3yz.2x 2 - 3xz - 2xy + 3yz= (2x 2 - 3xz) + (-2xy + 3yz)= x(2x - 3z) - y(2x - 3z)= (x - y)(2x - 3z)Example 4 Solve x 2 = 5x. Check thesolutions.Write the equation so that it is of the formab = 0.x 2 = 5x Original equationx 2 - 5x = 0Subtract 5x from each side.x(x - 5) = 0 Factor using the GCF, x.x = 0 or x - 5 = 0 Zero Product Propertyx = 5 Solve the equation.The roots are 0 and 5. Check bysubstituting 0 and 5 for x in theoriginal equation.462 Chapter 8 Factoring

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