Holt Algebra 2 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Warm Up Factor each expression 1 3 x 6 y 2 a 2 b 2 3 x 1 x 3 4 ID: 1047493
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1. 3-4Factoring PolynomialsHolt Algebra 2Warm UpLesson PresentationLesson QuizHolt McDougal Algebra 2
2. Warm UpFactor each expression.1. 3x – 6y2. a2 – b2 3. (x – 1)(x + 3)4. (a + 1)(a2 + 1)x2 + 2x – 33(x – 2y) (a + b)(a – b)a3 + a2 + a + 1 Find each product.
3. Use the Factor Theorem to determine factors of a polynomial.Objectives
4. Essential QuestionHow do you determine whether a linear binomial is a factor of a polynomial?
5. Recall that if a number is divided by any of its factors, the remainder is 0. Likewise, if a polynomial is divided by any of its factors, the remainder is 0. The Remainder Theorem states that if a polynomial is divided by (x – a), the remainder is the value of the function at a. So, if (x – a) is a factor of P(x), then P(a) = 0.
6. Determine whether the given binomial is a factor of the polynomial P(x). Example 1: Determining Whether a Linear Binomial is a FactorA. (x + 1); (x2 – 3x + 1) Find P(–1) by synthetic substitution or division. 1 –3 1 –1 –1 15–44 P(–1) = 5P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1. B. (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–2) by synthetic substitution or division.3 6 0 –5 –10 –2–6301000–500P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10.
7. Check It Out! Example 1Determine whether the given binomial is a factor of the polynomial P(x). a. (x + 2); (4x2 – 2x + 5) Find P(–2) by synthetic substitution or division. 4 –2 5 –2 –8 425–1020 P(–2) = 25P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5. b. (x – 2); (x4 – 2x3 + 2x2 + x – 10) 1 –2 2 1 –10 22101040520P(2) = 0, so (x – 2) is a factor of P(x) = x4 – 2x3 + 2x2 + x – 10. Divide the polynomial by 3, then find P(2) by synthetic substitution or division.
8. You are already familiar with methods for factoring quadratic expressions. You can factor polynomials of higher degrees using many of the same methods you learned in previous lessons.
9. Factor: x3 – x2 – 25x + 25.Example 2: Factoring by GroupingGroup terms.(x3 – x2) + (–25x + 25) Factor common monomials from each group.x2(x – 1) – 25(x – 1) Factor out the common binomial (x – 1).(x – 1)(x2 – 25) Factor the difference of squares.(x – 1)(x – 5)(x + 5)
10. Example 2 ContinuedCheck Use the table feature of your calculator to compare the original expression and the factored form.The table shows that the original function and the factored form have the same function values.
11. Check It Out! Example 2aFactor: x3 – 2x2 – 9x + 18.Group terms.(x3 – 2x2) + (–9x + 18) Factor common monomials from each group.x2(x – 2) – 9(x – 2) Factor out the common binomial (x – 2).(x – 2)(x2 – 9) Factor the difference of squares.(x – 2)(x – 3)(x + 3)
12. Check It Out! Example 2a ContinuedCheck Use the table feature of your calculator to compare the original expression and the factored form.The table shows that the original function and the factored form have the same function values.
13. Check It Out! Example 2bFactor: 2x3 + x2 + 8x + 4.Group terms.(2x3 + x2) + (8x + 4) Factor common monomials from each group.x2(2x + 1) + 4(2x + 1) Factor out the common binomial (2x + 1).(2x + 1)(x2 + 4) (2x + 1)(x2 + 4)
14. Essential QuestionHow do you determine whether a linear binomial is a factor of a polynomial?Use synthetic division or substitution. If the remainder is 0, then the linear binomial is a factor of the polynomial.
15. Factor the sum and difference of two cubes.Objectives
16. Essential QuestionsHow do you factor the sum of two cubes? How do you factor the difference of two cubes?
17. Just as there is a special rule for factoring the difference of two squares, there are special rules for factoring the sum or difference of two cubes.
18. Example 3A: Factoring the Sum or Difference of Two CubesFactor the expression.4x4 + 108xFactor out the GCF, 4x.4x(x3 + 27) Rewrite as the sum of cubes.4x(x3 + 33)Use the rule a3 + b3 = (a + b) (a2 – ab + b2).4x(x + 3)(x2 – x 3 + 32) 4x(x + 3)(x2 – 3x + 9)
19. Example 3B: Factoring the Sum or Difference of Two CubesFactor the expression.125d3 – 8Rewrite as the difference of cubes.(5d)3 – 23(5d – 2)[(5d)2 + 5d 2 + 22]Use the rule a3 – b3 = (a – b) (a2 + ab + b2).(5d – 2)(25d2 + 10d + 4)
20. Check It Out! Example 3aFactor the expression.8 + z6Rewrite as the sum of cubes.(2)3 + (z2)3(2 + z2)[(2)2 – 2 z² + (z2)2]Use the rule a3 + b3 = (a + b) (a2 – ab + b2).(2 + z2)(4 – 2z² + z4)
21. Check It Out! Example 3bFactor the expression.2x5 – 16x2Factor out the GCF, 2x2.2x2(x3 – 8) Rewrite as the difference of cubes.2x2(x3 – 23)Use the rule a3 – b3 = (a – b) (a2 + ab + b2).2x2(x – 2)(x2 + x 2 + 22) 2x2(x – 2)(x2 + 2x + 4)
22. Example 4: Geometry ApplicationThe volume of a plastic storage box is modeled by the function V(x) = x3 + 6x2 + 3x – 10. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).V(x) has three real zeros at x = –5, x = –2, and x = 1. If the model is accurate, the box will have no volume if x = –5, x = –2, or x = 1.
23. Use synthetic division to factor the polynomial. 1 6 3 –10 1 1 10V(x)= (x – 1)(x2 + 7x + 10) 107107Write V(x) as a product.V(x)= (x – 1)(x + 2)(x + 5) Factor the quadratic.Example 4 ContinuedOne corresponding factor is (x – 1).
24. Check It Out! Example 4The volume of a rectangular prism is modeled by the function V(x) = x3 – 8x2 + 19x – 12, which is graphed below. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).V(x) has three real zeros at x = 1, x = 3, and x = 4. If the model is accurate, the box will have no volume if x = 1, x = 3, or x = 4.
25. Use synthetic division to factor the polynomial. 1 –8 19 –12 1 1 10V(x)= (x – 1)(x2 – 7x + 12) 12–712–7Write V(x) as a product.V(x)= (x – 1)(x – 3)(x – 4) Factor the quadratic.Check It Out! Example 4 ContinuedOne corresponding factor is (x – 1).
26. Essential QuestionsHow do you factor the sum of two cubes? How do you factor the difference of two cubes?
27. 4. x3 + 3x2 – 28x – 60Lesson Quiz2. x + 2; P(x) = x3 + 2x2 – x – 2 1. x – 1; P(x) = 3x2 – 2x + 5 8(2p – q)(4p2 + 2pq + q2) (x + 3)(x + 3)(x – 3)3. x3 + 3x2 – 9x – 27P(1) ≠ 0, so x – 1 is not a factor of P(x). P(2) = 0, so x + 2 is a factor of P(x). 5. 64p3 – 8q3(x + 6)(x – 5)(x + 2)
28. Student: The artist Picasso must have been really good at algebra.Parent: Why do you say that?Student: He was a famous cubist so he probaby had to do a lot of factoring!