1 / 15

Section 6.6

Section 6.6. Finding Rational Zeros. Rational Zero Theorem Synthetic & Long Division Using Technology to Approximate Zeros. Today you will look at finding zeros of higher degree polynomials using the rational zero theorem, synthetic division, and technology. Rational Zero Theorem.

kemp
Download Presentation

Section 6.6

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 6.6 Finding Rational Zeros

  2. Rational Zero TheoremSynthetic & Long DivisionUsing Technology to Approximate Zeros Today you will look at finding zeros of higher degree polynomials using the rational zero theorem, synthetic division, and technology.

  3. Rational Zero Theorem If f(x) = anxn + . . . + a1x + a0 has integer coefficients, then every rational zero of f has the form: p factor of constant term a0 (c) --- = ----------------------------------------- q factor of leading coefficient an (a)

  4. EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f(x)=15x3+ 14x2 - 3x – 2. SolutionThe constant term is –2 and the leading coefficient is 15. Divide 1 and 2 by 1. Divide 1 and 2 by 3. Divide 1 and 2 by 5. Divide 1 and 2 by 15. There are 16 possible rational zeros.

  5. EXAMPLE: Solving a PolynomialEquation Solve: x4-6x2- 8x + 24 = 0. SolutionBecause we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots.

  6. EXAMPLE: Solving a PolynomialEquation 2 • 0 -6 -8 24 • 24-4-24 • 1 2 -2 -12 0 x-intercept: 2 Solve: x4-6x2- 8x + 24 = 0. SolutionThe graph of f(x) =x4-6x2- 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation. The zero remainder indicates that 2 is a root of x4-6x2- 8x + 24 = 0.

  7. EXAMPLE: Solving a PolynomialEquation (x – 2)(x3+ 2x2-2x- 12) = 0 This is the result obtainedfrom the synthetic division. Solve: x4-6x2- 8x + 24 = 0. SolutionNow we can rewrite the given equation in factored form. x4-6x2+ 8x + 24 = 0 This is the given equation. x – 2 = 0 or x3+ 2x2-2x- 12 = 0Set each factor equal to zero. Now we must continue by factoring x3 + 2x2 - 2x - 12 = 0

  8. EXAMPLE: Solving a PolynomialEquation These are the coefficients ofx3+ 2x2-2x- 12 = 0. • 1 2 -2 -12 • 2 8 12 • 1 4 6 0 The zero remainder indicates that 2 is a root ofx3+ 2x2-2x- 12 = 0. x-intercept: 2 Solve: x4-6x2- 8x + 24 = 0. SolutionBecause the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x3+ 2x2-2x- 12 = 0.

  9. EXAMPLE: Solving a PolynomialEquation (x – 2)(x3+ 2x2-2x- 12) = 0 This was obtainedfrom the first synthetic division. (x – 2)(x – 2)(x2+4x+ 6) = 0 This was obtainedfrom the second synthetic division. Solve: x4-6x2- 8x + 24 = 0. SolutionNow we can solve the original equation as follows. x4-6x2+ 8x + 24 = 0 This is the given equation. x – 2 = 0 or x – 2 = 0 or x2+4x+ 6 = 0Set each factor equal to zero. x= 2 x= 2 x2+4x+ 6 = 0Solve.

  10. EXAMPLE: Solving a PolynomialEquation We use the quadratic formula because x2+4x+ 6 = 0 cannot be factored. Let a= 1, b= 4, and c= 6. Multiply and subtract under the radical. Simplify. The solution set of the original equation is {2, -2 - i -2 + i }. Solve: x4-6x2- 8x + 24 = 0. SolutionWe can use the quadratic formula to solve x2+4x+ 6 = 0.

  11. Properties of Polynomial Equations 1. Ifa polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots. 2. If a+bi is a root of a polynomial equation (b 0), then the non-real complex number a-bi is also a root. Non-real complex roots, if they exist, occur in conjugate pairs.

  12. Zeros of Polynomial Functions Using a Graphing Calculator Approximate the real zeros of 1. Enter equation into Y1 2. Graph the equation 3. Go to CALC: zero 4. Move cursor to left of the zero – hit enter 5. Move cursor to right of zero – hit enter 6. Hit Enter one more time to see the zero. 7. The coordinates of the zero appear. Repeat the process to find all other real zeros.

  13. Find all possible rational zeros of the functions below using the Rational Zero Theorem.

  14. f(x) = x4 + 2x2 - 24 • f(x) = x4 + 2x2 – 24 • Find real roots on calculator: -2, 2 • Use synthetic division: (x – 2) (x3 + 2x2 + 6x + 12) (x – 2)(x + 2)(x2 – 6) Set x2 – 6 = 0 and solve: Solutions: x = 2, x = -2,

  15. Assignment Section 6.6: page 362 – 363 # 15 – 42 (every 3rd), 48 – 57 (every 3rd)

More Related