More Zeroes of Polynomials In this lecture we look more carefully at - PowerPoint PPT Presentation

More Zeroes of Polynomials In this lecture we look more carefully at zeroes of polynomials. (Recall: a zero of a polynomial is sometimes called a “root”.) Our goal in the next few presentations is to set up a strategy for Elementary Functions attempting to find (if possible) all the zeroes of a given polynomial. We Part 2, Polynomials will assume, for this section, that our polynomial has coefficients which are Lecture 2.5a, The Rational Root Test integers. We will then set up some tests to run on the polynomial so that we can make some guesses at possible roots of the polynomial and begin to factor it. Dr. Ken W. Smith The Fundamental Theorem of Algebra tells us that a polynomial of degree Sam Houston State University n has n zeroes, if we include complex roots and if we count the multiplicity of the roots. 2013 We will be particularly interested in finding all the zeroes for various polynomials of small degree, n = 3 , n = 4 or maybe n = 5 . Smith (SHSU) Elementary Functions 2013 1 / 35 Smith (SHSU) Elementary Functions 2013 2 / 35 The Rational Root Test The Rational Root Test Consider the simple linear polynomial 3 x − 5 . It has one zero, x = 5 3 . This zero, 5 3 , is a rational number with numerator given by the constant term 5 and denominator given by the leading coefficient 3 of this (small) A rational number is a number which can be written as a ratio b d where polynomial. both the numerator b and the denominator d are integers (whole numbers). This concept generalizes. If we are factoring a polynomial In this part of our lecture, we describe the set of all possible rational f ( x ) = a n x n + a n − 1 x n − 1 + ... + a 2 x 2 + a 1 x + a 0 numbers which might be the root of our polynomial. then when we eventually write out the factoring We will call this set of all possible rational numbers the rational test set ; it will be a list of numbers to examine in our hunt for roots. f ( x ) = ( d 1 x − b 1 )( d 2 x − b 2 ) · · · ( d n x − b n ) the products of the coefficients d 1 d 2 · · · d n must equal the leading coefficient a n and the products of the constants b 1 b 2 · · · b n must equal the constant term a 0 . This leads to the Rational Root Test. Smith (SHSU) Elementary Functions 2013 3 / 35 Smith (SHSU) Elementary Functions 2013 4 / 35

The Rational Root Test Some Worked Examples on the Rational Root Test Find the set of all possible rational zeroes of the given function, as given by the Rational Root Theorem. 1 f ( x ) = 2 x 3 + 5 x 2 − 4 x − 3 2 f ( x ) = 3 x 3 − 4 x 2 + 5 . 3 f ( x ) = 6 x 6 + 5 x 2 + x − 35 . If x = b d is a rational number that is the root (zero) of the polynomial f ( x ) = a n x n + ... + a 1 x + a 0 then the numerator b is a factor of the Solutions. constant term a 0 and the denominator d is a factor of the leading 1 The set of rational zeroes of f ( x ) = 2 x 3 + 5 x 2 − 4 x − 3 is limited to coefficient a n . fractions whose numerator divides 3 and whose denominator divides 2: Rational Test Set = {± 1 , ± 3 , ± 1 2 , ± 3 2 } . The effect of the Rational Root Test is that given a polynomial f ( x ) we 2 The set of rational zeroes of f ( x ) = 3 x 3 − 4 x 2 + 5 is limited to can create a “Test Set” of rational numbers to try as zeroes. fractions whose numerator divides 5 and whose denominator divides 3: Rational Test Set = {± 1 , ± 5 , ± 1 3 , ± 5 3 } . 3 The set of rational zeroes of f ( x ) = 6 x 6 + 5 x 2 + x − 35 is limited to fractions whose numerator divides 35 and whose denominator divides 6: Rational Test Set = {± 1 , ± 5 , ± 7 , ± 35 , ± 1 2 , ± 5 2 , ± 7 2 , ± 35 2 , ± 1 3 , ± 5 3 , ± 7 3 , ± 35 3 , ± 1 6 , ± 5 6 , ± 7 6 , ± 35 Smith (SHSU) Elementary Functions 2013 5 / 35 Smith (SHSU) Elementary Functions 2013 6 / 35 6 Zeroes of Polynomials Elementary Functions In the next presentation we will work through factoring a fifth degree Part 2, Polynomials polynomial and discover upper and lower bounds on the possible zeroes of Lecture 2.5b, Bounds on the Set of Zeroes a polynomial. Dr. Ken W. Smith (END) Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 7 / 35 Smith (SHSU) Elementary Functions 2013 8 / 35

Bounds to the set of zeroes Bounds on zeroes In this presentation we work through the details of trying to compute We are trying to factor (exactly) the zeroes of a polynomials. These techniques, over three f ( x ) = 2 x 5 − 3 x 4 + 14 x 3 + 15 x 2 − 34 x − 30 centuries old, are now aided by tools such as graphing calculators. with Rational Test Set equal to We work though an example in detail. Suppose we wish to factor {± 1 2 , ± 1 , ± 3 2 , ± 2 , ± 5 2 , ± 3 , ± 5 , ± 6 , ± 15 2 , ± 10 , ± 15 , ± 30 } . completely the polynomial We might begin by trying the easier numbers, the integers. Let us first f ( x ) = 2 x 5 − 3 x 4 + 14 x 3 + 15 x 2 − 34 x − 30 . divide f ( x ) by x − 1 , using synthetic division with c = 1 . 2 − 3 14 15 − 34 − 30 We first create a “test set” of rational roots to try. Since the constant term 30 has 1 , 2 , 3 , 5 , 6 , 10 , 15 , 30 as factors and the leading coefficient 2 1 2 − 1 13 28 − 6 has factors 1 and 2 then by the Rational Root Test, our test set of possible 2 − 1 13 28 − 6 − 36 rational roots is Rational Test Set = {± 1 2 , ± 1 , ± 3 2 , ± 2 , ± 5 2 , ± 3 , ± 5 , ± 6 , ± 15 So f (1) = − 36 and so x = 1 is not a zero. 2 , ± 10 , ± 15 , ± 30 } . This might be discouraging, but doing synthetic division with c = 1 was pretty easy! This is a large set of rational numbers to try! Smith (SHSU) Elementary Functions 2013 9 / 35 Smith (SHSU) Elementary Functions 2013 10 / 35 Bounds on zeroes Bounds on zeroes We are factoring We are factoring f ( x ) = 2 x 5 − 3 x 4 + 14 x 3 + 15 x 2 − 34 x − 30 f ( x ) = 2 x 5 − 3 x 4 + 14 x 3 + 15 x 2 − 34 x − 30 with Rational Test Set equal to with Rational Test Set = {± 1 2 , ± 1 , ± 3 2 , ± 2 , ± 5 2 , ± 3 , ± 5 , ± 6 , ± 15 2 , ± 10 , ± 15 , ± 30 } . {± 1 2 , ± 1 , ± 3 2 , ± 2 , ± 5 2 , ± 3 , ± 5 , ± 6 , ± 15 2 , ± 10 , ± 15 , ± 30 } . We tried c = 1 and got f (1) = − 36 . Now let’s try c = 2 . We tried c = 1 and got f (1) = − 36 and then tried c = 2 and got 2 − 3 14 15 − 34 − 30 2 − 3 14 15 − 34 − 30 2 4 2 32 94 120 2 4 2 32 94 120 2 1 16 47 60 90 2 1 16 47 60 90 So x = 2 is not a zero of f ( x ) . Frustrating! But notice two things here. We know then that since f (2) = 90 , there is a zero between x = 1 and First notice that the remainder is positive; f (2) = 90 . x = 2 . But notice also that the bottom row in our synthetic division with In our earlier work, we discovered that f (1) = − 36 and so, by the IVT, the c = 2 is all positive numbers. graph of the function f ( x ) crosses the x -axis between x = 1 and x = 2 ! We can conclude from our understanding of synthetic division that if we Since f (1) is negative and f (2) is positive then there is a zero somewhere were to try a larger positive number c greater than c = 2 then the between 1 and 2 ! This is important information! numbers on the bottom row would get even larger still and so there is no Smith (SHSU) Elementary Functions 2013 11 / 35 Smith (SHSU) Elementary Functions 2013 12 / 35 chance of a zero to the right of x = 2 . We have found an upper bound on