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The Rational Zero Theorem

The Rational Zero Theorem

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The Rational Zero Theorem

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  1. The Rational Zero Theorem The Rational Zero 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 Zero 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.

  2. 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. 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.

  3. 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.

  4. 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.

  5. 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

  6. 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.

  7. 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.

  8. 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.

  9. 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.

  10. Descartes' Rule of Signs Iff(x)=anxn+ an-1xn-1+… +a2x2+a1x + a0 be a polynomial with real coefficients. 1. The number of positive real zeros of f iseither equal to the number of sign changes of f(x)or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero. 2. The number of negative real zeros of f iseither equal to the number of sign changes of f(-x) or is less than that number by an even integer. If f(-x)has only one variation in sign, then f has exactly one negative real zero.

  11. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f(x).Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f(-x). We obtain this equation by replacing x with -x in the given function. f(x)= x3+ 2x2 + 5x + 4This is the given polynomial function. Replace xwith -x. f(-x)= (-x)3+2(-x)2+ 5(-x) + 4 =-x3+ 2x2 - 5x + 4 EXAMPLE:Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4.

  12. 1 2 3 EXAMPLE:Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f(x)= x3+ 2x2 + 5x + 4. Solution Now count the sign changes. f(-x)=-x3+ 2x2 - 5x + 4 There are three variations in sign. # of negative real zeros of fis either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 - 2 = 1 negative real zero.