The Rational Zero Theorem. The Rational Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives.

The Rational Zero Theorem

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-1x

n-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.

The Rational Zero Theorem

If f (x) anxn an-1x

n-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.

pq

pq

EXAMPLE 1: 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.

1 2 1 2 1 25 53 3 15 15

Factors of the constant term, 2Possible rational zerosFactors of the leading coefficient, 15

1, 21, 3, 5, 15

1, 2, , , , , ,

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 2: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.Solution Listing all possible rational roots, we have:

Factors of the constant term, 24Possible rational zerosFactors of the leading coefficient, 1

1, 2 3, 4, 6, 8, 12, 241

1, 2 3, 4, 6, 8, 12, 24

EXAMPLE 2: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.Solution We begin by listing all possible rational roots.

Factors of the constant term, 24Possible rational zerosFactors of the leading coefficient, 1

1, 2 3, 4, 6, 8, 12, 241

1, 2 3, 4, 6, 8, 12, 24

EXAMPLE 2: 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

The zero remainder indicates that 2 is a root of x4 6x2 8x + 24 0.

2 1 0 6 8 24 2 4 4 241 2 2 12 0

EXAMPLE 2: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.

Solution Now we can rewrite the given equation in factored form.

(x – 2)(x3 2x2 2x 12) 0 This is the result obtained from the synthetic division.

x – 2 0 or x3 2x2 2x 12 Set each factor equal to zero.

x4 6x2 8x + 24 0 This is the given equation.

Now we must continue by factoring

x3 + 2x2 - 2x - 12 = 0

EXAMPLE 2: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.

Solution 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.

x-intercept

: 2

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 2: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.

Solution Now we can solve the original equation as follows.

(x – 2)(x3 2x2 2x 12) 0 This was obtained from the first synthetic division.

x4 6x2 8x + 24 0 This is the given equation.

(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 Set each factor equal to zero.

x 2 x 2 x2 4x 6 Solve.

EXAMPLE 2: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.

Solution We can use the quadratic formula to solve x2 4x 6

Let a 1, b 4, and c 6.

24 4 4 1 62 1

We use the quadratic formula because x2 4x 6 cannot be factored.

2 42

b b acxa

Simplify.2 2i

Multiply and subtract under the radical. 4 82

4 2 22

i 8 4(2)( 1) 2 2i

The solution set of the original equation is {2, 2 i 2 i }.2,i 2i

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 non-real complex number a bi is also a root. Non-real complex roots, if they exist, occur in conjugate pairs.

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 non-real complex number a bi is also a root. Non-real complex roots, if they exist, occur in conjugate pairs.

Properties of Polynomial Equations

If f (x) anxn an1x

n1 … 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.

If f (x) anxn an1x

n1 … 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.

Descartes' Rule of Signs

EXAMPLE: Using Descartes’ Rule of Signs

Determine the possible number of positive and negative real zeros of f (x) x3 2x2 5x + 4.

Solution1. 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 x4

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

Determine the possible number of positive and negative real zeros of f (x) x3 2x2 5x + 4.

SolutionNow count the sign changes.

There are three variations in sign. # of negative real zeros of f is 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.

f (x) x3 2x2 5x + 4

1 2 3