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Bell Ringer

Learn about the Rational Root Theorem, exploring possible rational zeros of polynomial functions, solving equations, and analyzing polynomial end behavior. Examples and step-by-step solutions provided.

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Bell Ringer

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  1. Bell Ringer • 1. What is the Rational Root Theorem (aka Rational Zero Theorem Algebra II Textbook p. 378). • 2. What is the Fundamental Theorem of Algebra (search your notebook…Unit 2). • 3. If a function has 5 roots and only 1 of them is rational, what are the possibilities for the nature and number of the other 4 roots? • 4. How does end behavior work for both odd and even degree polynomial functions?

  2. Applying the Rational Root Theorem Thursday, March 17, 2016

  3. The Rational Root Theorem The Rational Root Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation. Not every number in the list will be a zero of the function, but every rational zero of the polynomial function will appear somewhere in the list. The Rational Root Theorem Iff(x)=anxn+ an-1xn-1+…+a1x + a0 has integer coefficients and (where is reduced) is a rational zero, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an.

  4. EXAMPLE: Using the Rational Root 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. The actual solution set to f(x)=15x3+ 14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions.

  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. Practice Classwork: Polynomial End Behavior Homework: Rational Root Theorem: List ALL possible roots based on the theorem, then find ALL the actual roots.

  13. Exit Ticket • 1. Explain the Rational Root Theorem • 2. Once you have the rational roots, how do you determine the rest of the roots?

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