1 / 13

Section 6.4

Section 6.4. Solving Polynomial Equations. y = 3 x + 1 y = –2 x + 6. –2x + 3 y = 0 x + 3 y = 3. Solving Polynomial Equations. ALGEBRA 2 LESSON 6-4. (For help, go to Lessons 6-2 and 6-3.). Graph each system. Find any points of intersection. 1. 2.

nusa
Download Presentation

Section 6.4

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Section 6.4 Solving Polynomial Equations

  2. y = 3x + 1 y = –2x + 6 –2x + 3y = 0 x + 3y = 3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (For help, go to Lessons 6-2 and 6-3.) Graph each system. Find any points of intersection. 1. 2. Factor each expression. 3.x2 – 2x – 15 4.x2 – 9x + 14 Check Skills You’ll Need 6-4

  3. y = 3x + 1 y = –2x + 6 –2x + 3y = 0 x + 3y = 3 2 3 y = x y = – x + 1 1 3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solutions 1.    2. intersection: (1, 4) Rewrite equations: intersection: (1, ) 3. Factors of –15 with a sum of –2: –5 and 3 x2 – 2x – 15 = (x – 5)(x + 3) 4. Factors of 14 with a sum of –9: –7 and –2 x2 – 9x + 14 = (x – 7)(x – 2) 2 3 6-4

  4. Step 1:  Graph y1 = x3 – 19x and y2 = –2x2 + 20 on a graphing calculator. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Graph and solve x3 – 19x = –2x2 + 20. Step 2:  Use the Intersect feature to find the x values at the points of intersection. The solutions are –5, –1, and 4. 6-4

  5. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (continued) Check: Show that each solution makes the original equation a true statement. x3 – 19x = –2x2 + 20 x3 – 19x = –2x2 + 20 (–5)3 – 19(–5) –2(–5)2 + 20  (–1)3 – 19(–1) –2(–1)2 + 20 –125 + 95 –50 + 20 –1 + 19 –2 + 20 –30 = –30 18 = 18 x3 – 19x = –2x2 + 20 (4)3 – 19(4) –2(4)2 + 20 64 – 76 –32 + 20 –12 = –12 Quick Check 6-4

  6. 15.9 ft3 • = 27475.2 in.3Convert the volume to cubic inches. Graph y1 = 27475.2 and y2 = (x + 4)x(x – 3).    Use the Intersect option of the calculator. When y = 27475.2, x 30. So x – 3 27 and x + 4 34. 123 in.3 ft3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Quick Check The dimensions in inches of the cubicle area inside a doghouse can be expressed as width x, length x + 4, and height x – 3. The volume is 15.9 ft3. Find the dimensions of the doghouse. V = l • w • hWrite the formula for volume. 27475.2 = (x + 4)x(x – 3) Substitute. The dimensions of the doghouse are about 30 in. by 27 in. by 34 in. 6-4

  7. Sum and Difference of Cubes

  8. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor x3 – 125. x3 – 125 = (x)3 – (5)3Rewrite the expression as the difference of cubes. = (x – 5)(x2 + 5x + (5)2) Factor. = (x – 5)(x2 + 5x + 25) Simplify. Quick Check 6-4

  9. 8x3 + 125 = (2x)3 + (5)3Rewrite the expression as the difference of cubes. = (2x + 5)((2x)2 – 10x + (5)2) Factor. = (2x + 5)(4x2 – 10x + 25) Simplify. 5 2 Since 2x + 5 is a factor, x = – is a root. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solve 8x3 + 125 = 0. Find all complex roots. The quadratic expression 4x2 – 10x + 25 cannot be factored, so use the Quadratic Formula to solve the related quadratic equation 4x2 – 10x + 25 = 0. 6-4

  10. –b ± b2 – 4ac 2a x = Quadratic Formula =    Substitute 4 for a, –10 for b, and 25 for c. –(–10) ± (–10)2 – 4(4)(25) 2(4) – (–10) ± –300 8 = Use the Order of Operations. 10 ± 10i 3 8 = –1 = 1 = Simplify. 5 2 5 ± 5i 3 4 5 ± 5i 3 4 The solutions are – and . Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (continued) Quick Check 6-4

  11. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor x4 – 6x2 – 27. Step 1:  Since x4 – 6x2 – 27 has the form of a quadratic expression, you can factor it like one. Make a temporary substitution of variables. x4 – 6x2 – 27 = (x2)2 – 6(x2) – 27 Rewrite in the form of a quadratic expression. = a2 – 6a – 27 Substitute a for x2. Step 2:  Factor a2 – 6a – 27. a2 – 6a – 27 = (a + 3)(a – 9) Step 3:  Substitute back to the original variables. (a + 3)(a – 9) = (x2 + 3)(x2 – 9) Substitute x2 for a. = (x2 + 3)(x + 3)(x – 3) Factor completely. Quick Check The factored form of x4 – 6x2 – 27 is (x2 + 3)(x + 3)(x – 3). 6-4

  12. x = ± 3 or x = ± i 5 Solve for x, and simplify. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solve x4 – 4x2 – 45 = 0. x4 – 4x2 – 45 = 0 (x2)2 – 4(x2) – 45 = 0    Write in the form of a quadratic expression. Think of the expression as a2 – 4a – 45, which factors as (a – 9)(a + 5). (x2 – 9)(x2 + 5) = 0 (x – 3)(x + 3)(x2 + 5) = 0 x = 3 or x = –3 or x2 = –5 Use the factor theorem. Quick Check 6-4

  13. ±2, ±i 7 5 ± 5i 3 4 –5, Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 1. Solve x3 – 2x2 – 3 = x – 4 by graphing. Where necessary, round to the nearest hundredth. Factor each expression. 2. 216x3 – 1 3. 8x3 + 125 4.x4 – 5x2 + 4 Solve each equation. 5.x3 + 125 = 0 6.x4 + 3x2 – 28 = 0 –0.80, 0.55, 2.25 (6x – 1)(36x2 + 6x + 1) (2x + 5)(4x2 – 10x + 25) (x + 1)(x – 1)(x + 2)(x – 2) 6-4

More Related