1. 6-4 Solving Polynomial Equations

2. Objectives Solving Equations by Graphing Solving Equations by Factoring

3. Graph and solve x 3 – 19x = –2x 2 + 20. Step 2: Use the Intersect feature to find the x values at the points of intersection. Step 1: Graph y 1 = x 3 – 19x and y 2 = –2x 2 + 20 on a graphing calculator. The solutions are –5, –1, and 4. Solving by Graphing

4. (continued) Check: Show that each solution makes the original equation a true statement. x 3 – 19x= –2x 2 + 20 x 3 – 19x = –2x 2 + 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 x 3 – 19x= –2x 2 + 20 (4) 3 – 19(4)–2(4) 2 + 20 64 – 76–32 + 20 –12= –12 Continued

5. 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 ft 3. Find the dimensions of the doghouse. V = l w hWrite the formula for volume. 27475.2 = (x + 4)x(x – 3)Substitute. 15.9 ft 3 = 27475.2 in. 3 Convert the volume to cubic inches. 12 3 in. 3 ft 3 The dimensions of the doghouse are about 30 in. by 27 in. by 34 in. Graph y 1 = 27475.2 and y 2 = (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. Real World Example

6. Vocabulary Sum of Cubes Difference of Cubes

7. Factor x 3 – 125. x 3 – 125 = (x) 3 – (5) 3 Rewrite the expression as the difference of cubes. = (x – 5)(x 2 + 5x + (5) 2 )Factor. = (x – 5)(x 2 + 5x + 25)Simplify. Factoring a Sum or Difference of Cubes

8. Solve 8x 3 + 125 = 0. Find all complex roots. 8x 3 + 125 = (2x) 3 + (5) 3 Rewrite the expression as the difference of cubes. = (2x + 5)((2x) 2 – 10x + (5) 2 )Factor. = (2x + 5)(4x 2 – 10x + 25)Simplify. The quadratic expression 4x 2 – 10x + 25 cannot be factored, so use the Quadratic Formula to solve the related quadratic equation 4x 2 – 10x + 25 = 0. Since 2x + 5 is a factor, x = – is a root. 5252 Solving a Polynomial Equation

9. (continued) x = Quadratic Formula = Substitute 4 for a, –10 for b, and 25 for c. –b ± b 2 – 4ac 2a –(–10) ± (–10) 2 – 4(4)(25) 2(4) = Use the Order of Operations. – (–10) ± –300 8 = –1 = 1 10 ± 10i 3 8 = Simplify. 5 ± 5i 3 4 The solutions are – and. 5252 5 ± 5i 3 4 Continued

10. Factor x 4 – 6x 2 – 27. Step 1: Since x 4 – 6x 2 – 27 has the form of a quadratic expression, you can factor it like one. Make a temporary substitution of variables. x 4 – 6x 2 – 27 = (x 2 ) 2 – 6(x 2 ) – 27Rewrite in the form of a quadratic expression. = a 2 – 6a – 27Substitute a for x 2. Step 2: Factor a 2 – 6a – 27. a 2 – 6a – 27 = (a + 3)(a – 9) Step 3: Substitute back to the original variables. (a + 3)(a – 9) = (x 2 + 3)(x 2 – 9)Substitute x 2 for a. = (x 2 + 3)(x + 3)(x – 3)Factor completely. The factored form of x 4 – 6x 2 – 27 is (x 2 + 3)(x + 3)(x – 3). Factoring by Using a Quadratic Pattern

11. Solve x 4 – 4x 2 – 45 = 0. x 4 – 4x 2 – 45 = 0 (x 2 ) 2 – 4(x 2 ) – 45 = 0 Write in the form of a quadratic expression. Think of the expression as a 2 – 4a – 45, which factors as (a – 9)(a + 5). (x 2 – 9)(x 2 + 5) = 0 (x – 3)(x + 3)(x 2 + 5) = 0 x = 3 or x = –3 or x 2 = –5Use the factor theorem. x = ± 3 or x = ± i 5Solve for x, and simplify. Solving a Higher-Degree a Polynomial Equation

12. Homework Pg 330 & 331 #1, 2,15, 21, 27