Polynomial Functions and EquationsContinuation

A polynomial function is a function of the form:

on

nn

n axaxaxaxf 1

11

All of these coefficients are real numbers

n must be a positive integer

The degree of the polynomial is the largest power on any x term in the polynomial.

an ≠ 0

Polynomial Functions

Given f(x) =an xn + an-1 xn-1+…+ a1 x + a0,

where an ≠ 0 and r is a real number,

the following statements are equivalent:

• x = r is a zero of the function f.

• x = r is a solution of the polynomial equation f(x) = 0.

• (x – r ) is a factor of the polynomial f(x).

• (r , 0) is an x-intercept of the graph of f.

RATIONAL ZEROS OF POLYNOMIAL FUNCTIONS

Rational Zeros of Polynomial Functions

The zeros of a polynomial function are the solutions that can be found when each of the factors of the polynomial is set equal to zero and the value of the variable is solved.

Fundamental Theorem of Algebra

Every polynomial function whose defining equation is f(x) =an xn + an-1 xn-1+…+ a1 x + a0, where an ≠ 0 and n ≥ 1 has at least one complex

zero.

Fundamental Theorem of Algebra

Every polynomial function whose defining equation is f(x) =an xn + an-1 xn-1+…+ a1 x + a0, where an ≠ 0 and n ≥ 1 has at least one complex zero.

Note that a complex zero of the polynomial can either be

a real or imaginary number.

Rational Zeros of Polynomial Functions

The factored form of the polynomialf(x) =an xn + an-1 xn-1+…+ a1 x + a0

isf(x) =an (x – r1) (x – r2)… (x – rn)

where r1, r2, … rn are the zeros of the polynomial.

If a factor (x – r) occurs k times, then r is called a zero of multiplicity k.

Example 1a.

Find the zeros of the given function and determine the multiplicity of each zero.

Solution: -1 is a zero of multiplicity 3 and 1 is a zero of multiplicity 1.

)1()1()( 3 xxxf

Example 1b.

Find the zeros of the given function and determine the multiplicity of each zero.

Solution: 1 is a zero of multiplicity 2, -3 is a zero of multiplicity 3, and 2 is a zero of multiplicity 1.

)2()3()1()( 32 xxxxf

Example 1c.Find the zeros of the given function and determine the multiplicity of each zero.

Solution: ½ and -½ are zeros of multiplicity 1, 3/2 is a zero of multiplicity 4, and -3 is a zero with multiplicity 5.

54 )3()32)(12)(12()( xxxxxf

Example 2a.Find a polynomial equation of least possible degree which satisfies the given condition: zeros are 2, 1, and ¼

Solution:If f(r ) = 0, then (x – r) is a factor of f(x). Therefore, the desired polynomial is f(x) = (x – 2)(x – 1)( x – ¼).

Example 2b.Find a polynomial equation of least possible degree which satisfies the given condition: zeros are -1, 3 and 1 of multiplicity 2

Solution:If f(r ) = 0, then (x – r) is a factor of f(x). Therefore, the desired polynomial is f(x) = (x +1)(x – 3)(x – 1)2

Example 2c.Find a polynomial equation of least possible degree which satisfies the given condition: f(1) = f(2) = f(3) = f(0) = 0, f(4) = 24

Solution:By the Factor Theorem, f(x) = an (x – r1) (x – r2)… (x – rn)Since f(1) = f(2) = f(3) = f(0) = 0, then (x – 0),

(x – 1), (x – 2), and (x – 3) are factors of the polynomial f(x). If f(4) = 24, then

24 = an (4 – 0)(4 – 1)(4 – 2)(4 – 3) Therefore, the desired polynomial is f(x) = x(x – 1)(x – 2)(x – 3)

Descartes’s Rule of SignsIf f(x) =an xn + an-1 xn-1+…+ a1 x + a0 is a polynomial with real number coefficients, then the following holds true: 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.

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 3a.Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:

Solution:Apply the Descartes’s Rule of SignsSince f(x) is a polynomial of 3rd , there are 3 zeros of the function.There is no variation of signs in f(x) so there no positive roots.This implies that all roots are negative.Now, There are 3 variations of signs in f(x) so there are 3 negative roots.Therefore, all 3 roots of the function are negative.

133)( 23 xxxxf

133)( 23 xxxxf

Example 3b.Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:

Solution:Apply the Descartes’s Rule of SignsSince f(x) is a polynomial of 3rd degree, there are 3 zeros of the function.There is only 1 variation of signs in f(x) so there are either 1 positive root or none at all.Now, There are 2 variations of signs in f(x) so there are either 2 negative roots or none at all.Therefore, there are 1 positive and 2 negative roots of the function.

1)( 23 xxxxf

1)( 23 xxxxf

Example 3c.Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:

Solution:Apply the Descartes’s Rule of SignsSince f(x) is a polynomial of 4th degree, there are 4 zeros of the function.There are 2 variations of signs in f(x) so there are either 2 positive roots or none at all.Now, There are 2 variations of signs in f(x) so there are either 2 negative roots or none at all.Therefore, there are 2 positive and 2 negative roots of the function.

45)( 24 xxxf

45)( 24 xxxf

Example 3d.Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:

Solution:Apply the Descartes’s Rule of SignsSince f(x) is a polynomial of 3rd degree, there are 3 zeros of the function.There are 2 variation of signs in f(x) so there are either 2 positive roots or none at all.Now, . There is only 1 variation of sign in f(x) so there is exactly 1 negative root.Therefore, there are 2 positive roots and 1 negative root.

15411612)( 23 xxxxf

15411612)( 23 xxxxf

Example 3e.Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:

Solution:Apply the Descartes’s Rule of SignsSince f(x) is a polynomial of 5th degree, there are 5 zeros of the function.There are 5 variations of signs in f(x) so there are either 5 or 3 positive roots.Now, There is no variation of sign in f(x) so there is no negative root.Therefore, there are 5 positive roots of the function.

1202742258515)( 2345 xxxxxxf

1202742258515)( 2345 xxxxxxf

The Rational Zero Theorem

If the rational number p/q , a fraction in lowest terms, is a root of the equation

f(x) =an xn + an-1 xn-1+…+ a1 x + a0,

where each a1 , a2 ,…, an is an integral coefficient, then p is an exact divisor of an and q is an exact divisor of a0 .

Example 4a.Determine the zeros of the given polynomial function:

Solution:Applying the Descartes’s Rule of Signs (see Example 3a), there are 3 negative roots. By the Rational Zero Theorem, p must be an exact divisor of 1 ; p = 1, -1 q must be an exact divisor of 1; q = 1, -1 the set of possible rational zeros p/q of f(x) is {1, -1}.

However, since there are 3 negative roots, we will only try x = -1. Perform synthetic division.

133)( 23 xxxxf

1 3 3 1 -1

1

-1

2

-2

1

-1

0

1 2 1 -1

1

-1

1

-1

1133 23

xxxx

11 xx

)1)(12(133 223 xxxxxx

1122

xxx

11 xx

0

323

23

)1(133

)1)(1)(1(133

xxxx

xxxxxx

122 xx

Therefore, the zero of f(x) is 1 of multiplicity 3.

Example 4b.Determine the zeros of the given polynomial function:

Solution:Applying the Descartes’s Rule of Signs (see Example 3b), there are 1 positive and 2 negative roots. By the Rational Zero Theorem, p must be an exact divisor of -1 ; p = 1, -1 q must be an exact divisor of 1; q = 1, -1 the set of possible rational zeros p/q of f(x) is {1, -

1}.

1)( 23 xxxxf

Example 4b.Determine the zeros of the given polynomial function:

Solution:Now use the values in the set of possible rational zeros to find the roots of the function.Suppose x = 1. Perform synthetic division.

1)( 23 xxxxf

1 1 -1 -1 1

1

1

2

2

1

1

0

1 2 1 -1

1

-1

1

-1

1123

xxxx

11 xx

)1)(12(1 223 xxxxxx

1122

xxx

11 xx

0

223

23

)1)(1(1

)1)(1)(1(1

xxxxx

xxxxxx

122 xx

Therefore, the zeros of f(x) are -1, of multiplicity 2, and 1.

Example 4c.Determine the zeros of the given polynomial function:

Solution:Applying the Descartes’s Rule of Signs, there are 2 positive and 2 negative roots. By the Rational Zero Theorem, p must be an exact divisor of 4 ; p = 1, -1, 2, -2, 4,

-4 q must be an exact divisor of 1; q = 1, -1 the set of possible rational zeros p/q of f(x) is {1, 2, 4}.

45)( 24 xxxf

1 0 -5 0 4 1

1

1

1

1

-4

-4

0

1 1 -4 -4 -1

1

-1

0

0

145 24

xxx

11 xx

)1)(44(45 2324 xxxxxx

14423

xxxx

11 xx

0

4423 xxx

So,

-4

-4

4

-4

Note that )2)(2()4( 2 xxx)1)(1)(2)(2(45 24 xxxxxx

)4( 2 x

Therefore, the zeros of f(x) are 2, -2, 1, and -1.

Exercises:1.Find the zeros of the given functions and determine the multiplicity of each zero.1.11.21.32. Find a polynomial equation of least possible degree which satisfies the given conditions: 2.1 zeros are 1, -1, and ¾2.2 Zeros are 4 of multiplicity 3, -1 of multiplicity 2, and 3

)12()5()( 2 xxxf)23)(1)(3()( xxxxf

32 )2()3()( xxxxf

Exercises:3. Determine the maximum number of positive roots and the maximum number of negative roots of the given polynomial function:3.13.23.34. Determine the zeros of the given polynomial functions:4.1 4.24.3

1)( 4 xxf364914)( 246 xxxxf

xxxxf 910)( 35

xxxxf 6)( 23 67)( 3 xxxf

2446296)( 23 xxxxf