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X 4 – 81 = (x 2 + 9) (x + 3) (x-3) Solving Polynomial Equations At the end of this two-part lesson, you will:  Solve Polynomial Equations by Factoring/Using.

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Presentation on theme: "X 4 – 81 = (x 2 + 9) (x + 3) (x-3) Solving Polynomial Equations At the end of this two-part lesson, you will:  Solve Polynomial Equations by Factoring/Using."— Presentation transcript:

1 X 4 – 81 = (x 2 + 9) (x + 3) (x-3) Solving Polynomial Equations At the end of this two-part lesson, you will:  Solve Polynomial Equations by Factoring/Using Quadratic Formula  Solve Polynomial Equations by Graphing X 3 – 125 = 0 (x-5) (x 2 +5x + 25) x = 5; x = ?

2 RECALL: Quadratic Formula Sums and Differences of Cubes u 3 + v 3 = ___________________________ u 3 - v 3 = ___________________________ x = ( u + v) (u 2 – uv + v 2 ) ( u - v) (u 2 + uv + v 2 ) NOTE: Equation must be in standard form ax 2 + bx + c = 0, where a, b, and c are and a > 0.

3 Review: Factoring Sums and Differences of Cubes Being familiar with perfect cubes will make factoring sums and differences much easier! x 3 – 512 x 3 + 125 x 3 – 343 x 3 + 216 x 3 – 8 3 = (x – 8) (x 2 + 8x + 64) x 3 – 7 3 = (x – 7) (x 2 + 7x + 49) x 3 + 5 3 = (x + 5) (x 2 - 5x + 25) x 3 + 6 3 = (x + 6) (x 2 - 6x + 36)

4 SOLVING A POLYNOMIAL EQUATION Solve 8x 3 + 125 = 0. Find all complex roots. You will need to use Quadratic Formula. ( ) 3 + __ 3 =

5 MENTAL MATH: FACTORING HIGHER-DEGREE POLYNOMIALS BY USING A QUADRATIC FORM x 4 – 3x 2 – 10x 4 + 11x 2 + 18 x 4 – 8x 2 + 12x 4 – 17x 2 + 72 x 4 – 5x 2 + 6x 4 – 16x 2 + 63 x 4 – 7x 2 + 12x 4 + 6x 2 + 72 Factor completely.

6 Examples Solving Higher-Degree Equations by Using a Quadratic Form Solve x 4 – 6x 2 -27 = 0. Solve x 4 – 3x 2 - 10 = 0. Solve x 4 + 11x2 +10 = 0. Solve x 4 – 4x2 - 45 = 0. Pick your poison!

7 ExampleSolving Higher-Degree Polynomial Equations by Using Factoring By Grouping 30X 3 + 40X 2 + 3X + 4 = 0 3X 4 + 3X 2 + 6X 2 + 6X = 0 X 4 + 12X 3 + 4X 2 + 48X = 0 Pick your poison!

8 Solving by Graphing You can also solve a polynomial equation by graphing each side of the equation separately, and finding the x-values (zeros) at the point(s) of intersection. y = x 3 + 3x 2 – x - 3 Step 1: Graph y 1 = x 3 + 3x 2 & y 2 = x + 3x Step 2: Use the intersect feature to find the x-values (zeros) of the points of intersection. The solutions are -3, 1 & 1. …on a final note!

9 FINAL CHECKS FOR UNDERSTANDING 1.What does it mean for a polynomial with integer coefficients to be completely factored with respect to the integers? 2. Give an example of a second-degree polynomial that cannot be factored with respect to the integers. 3.Factor completely with respect to the integers: 81x 4 – 1 16x 3 – 4 1 – 64x 3 2x 3 – 8x 2 + 3x - 12

10 Homework Assignment: Pages 324-325. 1-7 odd (calculator) 12-32 all. Column 1 (42-58 all), 61, 62. Reminder: Chapter 6 Exam on _____________________.


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