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Section 6.4

Section 6.4

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Section 6.4

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  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