The Rational Zero Theorem

1. The Rational Zero Theorem 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. 3.4: Zeros of Polynomial Functions 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.

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. 3.4: Zeros of Polynomial Functions EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f(x)=15x3+ 14x2 - 3x – 2.

3. EXAMPLE: Solving a PolynomialEquation 3.4: Zeros of Polynomial Functions Solve: x4-6x2- 8x + 24 = 0. SolutionRecall that we refer to the zeros of a polynomial function and the roots of a polynomial equation. Because 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 3.4: Zeros of Polynomial Functions 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. 3.4: Zeros of Polynomial Functions 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.

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 3.4: Zeros of Polynomial Functions Solve: x4-6x2- 8x + 24 = 0. SolutionWe can use the same approach to look for rational roots of the polynomial equation x3+ 2x2-2x- 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x4-6x2- 8x + 24 = 0. Because 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, confirmed by the following synthetic division.

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. 3.4: Zeros of Polynomial Functions 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 }. 3.4: Zeros of Polynomial Functions Solve: x4-6x2- 8x + 24 = 0. SolutionWe can use the quadratic formula to solve x2+4x+ 6 = 0.

9. 3.4: Zeros of Polynomial Functions 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 nonreal complex number a-bi is also a root. Nonreal complex roots, if they exist, occur in conjugate pairs.

10. 3.4: Zeros of Polynomial Functions 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 3.4: Zeros of Polynomial Functions 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 3.4: Zeros of Polynomial Functions 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. The number of negative real zeros of fis either equal to the number of sign changes, 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.