The Rational Zero Theorem

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The Rational Zero Theorem

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Presentation transcript:

The Rational Zero Theorem 3.4: Zeros of Polynomial Functions 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 If f (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.

EXAMPLE: Using the Rational Zero Theorem 3.4: Zeros of Polynomial Functions EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f (x) = 15x3 + 14x2 - 3x – 2. Solution The constant term is –2 and the leading coefficient is 15. Divide 1 and 2 by 1. by 3. by 5. 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.

EXAMPLE: Solving a Polynomial Equation 3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution Recall 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.

EXAMPLE: Solving a Polynomial Equation 3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution The 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. x-intercept: 2 2 0 -6 -8 24 2 4 -4 -24 1 2 -2 -12 0 The zero remainder indicates that 2 is a root of x4 - 6x2 - 8x + 24 = 0.

EXAMPLE: Solving a Polynomial Equation 3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution Now we can rewrite the given equation in factored form. x4 - 6x2 + 8x + 24 = 0 This is the given equation. (x – 2)(x3 + 2x2 - 2x - 12) = 0 This is the result obtained from the synthetic division. x – 2 = 0 or x3 + 2x2 - 2x - 12 = 0 Set each factor equal to zero.

EXAMPLE: Solving a Polynomial Equation 3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution We 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. x-intercept: 2 1 2 -2 -12 2 8 12 1 4 6 0 These are the coefficients of x3 + 2x2 - 2x - 12 = 0. The zero remainder indicates that 2 is a root of x3 + 2x2 - 2x - 12 = 0.

EXAMPLE: Solving a Polynomial Equation 3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution Now we can solve the original equation as follows. x4 - 6x2 + 8x + 24 = 0 This is the given equation. (x – 2)(x3 + 2x2 - 2x - 12) = 0 This was obtained from the first synthetic division. (x – 2)(x – 2)(x2 + 4x + 6) = 0 This was obtained from the second synthetic division. x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = 0 Set each factor equal to zero. x = 2 x = 2 x2 + 4x + 6 = 0 Solve.

EXAMPLE: Solving a Polynomial Equation 3.4: Zeros of Polynomial Functions EXAMPLE: Solving a Polynomial Equation Solve: x4 - 6x2 - 8x + 24 = 0. Solution We can use the quadratic formula to solve x2 + 4x + 6 = 0. 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 }.

Properties of Polynomial Equations 3.4: Zeros of Polynomial Functions Properties of Polynomial Equations 1. If a 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.

Descartes' Rule of Signs 3.4: Zeros of Polynomial Functions Descartes' Rule of Signs If f (x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial with real coefficients. 1. The number of positive real zeros of f is either 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 is either 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.

EXAMPLE: Using Descartes’ Rule of Signs 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 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) = (-x)3 + 2(-x)2 + 5(-x) + 4 f (x) = x3 + 2x2 + 5x + 4 This is the given polynomial function. Replace x with -x. = -x3 + 2x2 - 5x + 4

EXAMPLE: Using Descartes’ Rule of Signs 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 1 2 3 There are three variations in sign. The number of negative real zeros of f is 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.