Elementary Functions More Zeroes of Polynomials the Rational Root Test

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 2013 multiplicity of the roots. 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 3x − 5. 5 It has one zero, x = 3 . 5 This zero, 3 , is a rational number with numerator given by the constant term 5 and denominator given by the leading coeﬃcient 3 of this (small) b A rational number is a number which can be written as a ratio 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 n n−1 2 numbers which might be the root of our polynomial. f(x) = anx + an−1x + ... + a2x + a1x + a0 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) = (d1x − b1)(d2x − b2) ··· (dnx − bn)

the products of the coeﬃcients d1d2 ··· dn must equal the leading coeﬃcient an and the products of the constants b1b2 ··· bn must equal the constant term a0.

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) = 2x3 + 5x2 − 4x − 3 2 f(x) = 3x3 − 4x2 + 5. b 3 6 2 If x = d is a rational number that is the root (zero) of the polynomial f(x) = 6x + 5x + x − 35. f(x) = a xn + ... + a x + a then the numerator b is a factor of the n 1 0 Solutions. constant term a0 and the denominator d is a factor of the leading 1 The set of rational zeroes of f(x) =2 x3 + 5x2 − 4x − 3 is limited to coeﬃcient an. fractions whose numerator divides3 and whose denominator divides2: 1 3 The eﬀect of the Rational Root Test is that given a polynomial f(x) we Rational Test Set = {±1, ±3, ± 2 , ± 2 }. 3 2 can create a “Test Set” of rational numbers to try as zeroes. 2 The set of rational zeroes of f(x) =3 x − 4x +5 is limited to fractions whose numerator divides5 and whose denominator divides3: 1 5 Rational Test Set = {±1, ±5, ± 3 , ± 3 }. 3 The set of rational zeroes of f(x) =6 x6 + 5x2 + x − 35 is limited to fractions whose numerator divides 35 and whose denominator divides 6: Rational Test Set = 1 5 7 35 1 5 7 35 1 5 7 35 Smith (SHSU) Elementary Functions 2013 5 / 35 {±Smith1, ± (SHSU)5, ±7, ±35, ± 2 , ± 2Elementary, ± 2 , ± Functions2 , ± 3 , ± 3 , ± 3 , ± 3 , ± 6 , ±20136 , ± 6 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.

(END) Dr. Ken W. Smith

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 centuries old, are now aided by tools such as graphing calculators. f(x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 with Rational Test Set equal to We work though an example in detail. Suppose we wish to factor 1 3 5 15 {± 2 , ±1, ± 2 , ±2, ± 2 , ±3, ±5, ±6, ± 2 , ±10, ±15, ±30}. completely the polynomial We might begin by trying the easier numbers, the integers. Let us ﬁrst 5 4 3 2 f(x) = 2x − 3x + 14x + 15x − 34x − 30. divide f(x) by x − 1, using synthetic division with c = 1.

We ﬁrst create a “test set” of rational roots to try. Since the constant 2 − 3 14 15 − 34 − 30 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 1 3 5 15 So f(1) = −36 and so x = 1 is not a zero. Rational Test Set = {± , ±1, ± , ±2, ± , ±3, ±5, ±6, ± , ±10, ±15, ±30}. 2 2 2 2 This might be discouraging, but doing synthetic division with c = 1 was This is a large set of rational numbers to try! pretty easy! 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 5 4 3 2 f(x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 f(x) = 2x − 3x + 14x + 15x − 34x − 30 with Rational Test Set equal to with Rational Test Set = {± 1 , ±1, ± 3 , ±2, ± 5 , ±3, ±5, ±6, ± 15 , ±10, ±15, ±30}. 1 3 5 15 2 2 2 2 {± 2 , ±1, ± 2 , ±2, ± 2 , ±3, ±5, ±6, ± 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 chanceSmithof a (SHSU) zero to the right ofElementaryx = 2. FunctionsWe have found an upper bound2013 12on / 35 the zeroes of f(x). An Upper Bound An Upper Bound

We are factoring f(x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30

We tried c = 2 and synthetic division gave us Our Rational Test Set is 2 − 3 14 15 − 34 − 30 {± 1 , ±1, ± 3 , ±2, ± 5 , ±3, ±5, ±6, ± 15 , ±10, ±15, ±30}. 2 4 2 32 94 120 2 2 2 2 2 1 16 47 60 90 We found c = 2 is an upper bound on the zeroes of f(x) Notice that if we try a larger positive number c greater than c = 2 then 5 15 since the number in the middle row are created by multiplying by c, then This immediately rules out 2 , 3, 5, 6, 2 , 10, 15, 30 as possible zeroes We the numbers on the bottom row will get even larger than they are now. So need not try any of these. there is no chance of a zero to the right of x = 2. An upper bound for polynomial zeroes: If, upon doing synthetic division with a positive value c, the bottom row in our computation of f(c) consists of all positive numbers (or zero) then c is an upper bound for the zeroes of f(x). We shouldSmith (SHSU)not look for zeroes furtherElementary Functions to the right of c. 2013 13 / 35 Smith (SHSU) Elementary Functions 2013 14 / 35 An Upper Bound Finding zeroes

We continue to attempt to factor We discovered that f(x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x4 + 7x2 + 18x + 10) Let us go back to our observation that there is a zero between x = 1 and We want to find more roots of f(x) but since we have factored out a x = 2. This suggests that we try x = 3 as a root. We do the synthetic 2 linear term, let us now focus on factoring x4 + 7x2 + 18x + 10. division. 2 − 3 14 15 − 34 − 30 There is an important principle here: once we have found a factor, concentrate on the quotient that remains. Do not waste time by returning 3 3 0 21 54 30 2 to the original polynomial. 2 0 14 36 20 0 Is it clear that this new polynomial (x4 + 7x2 + 18x + 10) has no positive 3 Success!! So x = 2 is a root of f(x) and f(x) factors as zeroes? 3 If we try synthetic division with c = 0 we would just get, as bottom row, 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (x − )(2x4 + 14x2 + 36x + 20). 2 the coefficients 1, 0, 7, 18, 10 which are already positive. Anything to the It is probably better if we factor a 2 out of the right-hand factor and right of zero will only makes these numbers bigger. multiply it into the linear term and rewrite this as So we should try some negative numbers. 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x4 + 7x2 + 18x + 10) At this point, since no positive numbers could give a zero and since this Smith (SHSU) Elementary Functions 2013 15 / 35 Smith (SHSU) Elementary Functions 2013 16 / 35 polynomial has constant term 10 and leading coefficient 1, the Test Set of possible rational roots has shrunk to {−10, −5, −2, −1}. Finding zeroes Finding zeroes

We discovered that We discovered that 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x4 + 7x2 + 18x + 10) 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x + 1)(x3 − x2 + 8x + 10) and that there are no more positive roots to ﬁnd. and that there are no more positive roots to ﬁnd. 3 2 Let us try c = −1. We continue on with our factoring by trying to factor x − x + 8x + 10. Let’s try c = −2. 1 0 7 18 10 1 − 1 8 10 − 1 − 1 1 − 8 − 10 − 2 − 2 6 − 28 1 − 1 8 10 0 1 − 3 14 − 18

We have found another factor! So So f(−2) = −18 and so x = −2 is not a zero. Notice the pattern across the bottom row in our synthetic division. It x4 + 7x2 + 18x + 10 = (x + 1)(x3 − x2 + 8x + 10) alternates, positive 1, negative 3, positive 14, negative 18. If we were try a negative number to the left of x = −2 on the real line, it would make the and so negative 3 more negative, which in turn would give a larger positive value 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x + 1)(x3 − x2 + 8x + 10). to the next entry, leading to a bottom line entry larger than positive 14

Smith (SHSU) Elementary Functions 2013 17 / 35 and then,Smithin (SHSU) the next step, a numberElementary more Functions negative than negative2013 18. The 18 / 35 numbers at each stage are further from zero than they are here. A lower bound on zeroes A lower bound on zeroes

To illustrate this, here is the synthetic division with c = −3 and c = −4.

1 − 1 8 10 1 − 1 8 10 − 3 − 3 12 − 60 − 2 − 2 6 − 28 1 − 4 20 − 50 1 − 3 14 − 18

1 − 1 8 10 A lower bound for the zeroes of a polynomial: If, upon doing synthetic − 4 − 4 20 − 112 division with a negative value c, the bottom row in our computation of f(c) consists of numbers alternating in sign then c is an lower bound for 1 − 5 28 − 102 the zeroes of f(x).

Notice how the bottom rows continued to alternate, with larger and larger We should not look for zeroes further to the left of c on the number line. absolute value. So x = −2 is a lower bound for our possible roots; there is no reason to (For the purpose of this result, we can treat zero as positive or negative, try anything smaller. giving it whatever sign we wish.) We summarize what we have learned here by describing when we know we have a lower bound for our roots. Smith (SHSU) Elementary Functions 2013 19 / 35 Smith (SHSU) Elementary Functions 2013 20 / 35 Finding zeroes Finding zeroes

Let’s go back and look at our cubic g(x) = x3 − x2 + 8x + 10. It has y-intercept (0, 10). It is a cubic polynomial with end behavior .% so we know that although g(0) = 10, eventually to the left of x = 0 the Returning to our earlier factoring problem. function becomes negative. We discovered that By the IVT, this cubic polynomial has a root which is negative, which we 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x + 1)(x3 − x2 + 8x + 10). have not yet found. and that there are no more positive roots to ﬁnd, and that −2 is a lower Did we try everything? Almost. We tried x = −1, which was a zero of bound of the roots of this polynomial. f(x) and then we agreed that x = −2 was a lower bound on zeroes of f(x). We have now ruled out everything else is our Test Set, while discovering What we did not do is test x = −1 twice! Recall that a polynomial can that x = 3 and x = −1 are zeroes of our polynomial. Now what do we do? 2 have a zero with multiplicity two or more.... Let us test x = −1, using synthetic division, with the cubic x3 − x2 + 8x + 10. 1 − 1 8 10

Smith (SHSU) Elementary Functions 2013 21 / 35 Smith (SHSU) − 1 Elementary− Functions1 2 − 10 2013 22 / 35 Finding zeroes Finding zeroes 1 − 2 10 0 So x = −1 is a zero a second time and x + 1 is a factor, twice, of f(x). We now have So x2 − 2x + 10 factors into x2 − 2x + 10 = (x − (2 + 3i))(x − (2 − 3i)). 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30 = (2x − 3)(x + 1)2(x2 − 2x + 10) (1) Our ﬁnal factoring of the ﬁfth degree polynomial f(x) is then f(x) = 2x5 −3x4 +14x3 +15x2 −34x−30 = (2x−3)(x4 +7x2 +18x+10) Once we reach a quadratic polynomial, we are almost done. Factoring = (2x − 3)(x + 1)(x3 − x2 + 8x + 10) = (2x − 3)(x + 1)2(x2 − 2x + 10) quadratics are easy! = (2x − 3)(x + 1)2(x − (2 + 3i))(x − (2 − 3i)) We can use the quadratic formula if we don’t see an obvious factoring. In Here is the graph of y = f(x). this case, if x2 − 2x + 10 = 0 then √ 4 ± −36 4 ± 6i x = = = 2 ± 3i. 2 2

Thus the quadratic equation x2 − 2x + 10 has two complex roots (appearing, of course, as conjugate pairs.)

Smith (SHSU) Elementary Functions 2013 23 / 35 Smith (SHSU) Elementary Functions 2013 24 / 35 Finding Zeroes

Elementary Functions Part 2, Polynomials In the next lecture we review these upper and lower bound results and Lecture 2.5c, Bounding the Location of Zeroes then look at Descartes’ Rule of Signs.

(END) Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 25 / 35 Smith (SHSU) Elementary Functions 2013 26 / 35 Upper Bound Lower Bound

In a previous lecture we completely factored a ﬁfth degree polynomial, discovering along the way, some upper and lower bounds for the roots of a polynomial. If, upon doing synthetic division with a negative value c, the bottom row in our computation of f(c) consists of numbers alternating in sign then c We summarize what we learned about the upper and lower bounds for our is an lower bound for the zeroes of f(x). We should not look for zeroes set of real zeroes. further to the left of c on the number line. If, upon doing synthetic division with a positive value c, the bottom row in (For the purpose of this result, we can treat zero as positive or negative, our computation of f(c) consists of all positive numbers (or zero) then c is giving it whatever sign we wish.) an upper bound for the zeroes of f(x). We should not look for zeroes further to the right of c.

(For the purpose of this result, we can treat zero as positive.)

Smith (SHSU) Elementary Functions 2013 27 / 35 Smith (SHSU) Elementary Functions 2013 28 / 35 Descartes’ Rule of Signs Descartes’ Rule of Signs, Positive version

We have one more guide in our search for roots of a polynomial. It is a Examples. “rule” which is four centuries old, discovered by RenéDescartes. 1 The polynomial x3 − 8 has coefficients 1, (0, 0, ) − 8. Ignore the Descartes’ Rule of Signs (Positive version) zeroes; there is one change of sign, from 1 to -8. So the polynomial has 1 positive root. List the coefficients of a polynomial f(x), from leading coefficient to the 2 The polynomial f(x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30, studied constant term. Count the change of signs. This is an upper bound on the earlier, has coefficients 2, −3, 14, 15, −34, −30. This changes sign number of positive roots. 3 times (from 2 to −3, from −3 to 14 and from 15 to −34.) An upper bound for the number of positive roots of f(x) is 3. The polynomial The true number of positive roots may vary from this upper bound by a either has 3 positive roots or 1. (As we saw in our work, there was a multiple of two (since complex number occur in conjugate pairs.) pair of complex numbers, and so there was only one positive root.)

Smith (SHSU) Elementary Functions 2013 29 / 35 Smith (SHSU) Elementary Functions 2013 30 / 35 Descartes’ Rule of Signs, Negative version Descartes’ Rule of Signs, Negative version

Examples. A second version of Descartes’ Rule of Signs is... 1 Consider the polynomial g(x) = x3 − 8. g(−x) = −x3 − 8 has Given the polynomial f(x), list the coefficients of f(−x) (note the coefficients −1, (0, 0, ) − 8. Ignore the zeroes; there is no change of insertion of −x!), from leading coefficient to the constant term. Count the sign so the polynomial has no negative roots. change of signs. 2 Or consider the polynomial f(x) = 2x5 − 3x4 + 14x3 + 15x2 − 34x − 30, studied earlier. This is an upper bound on the number of negative roots. f(−x) = −2x5 − 3x4 − 14x3 + 15x2 + 34x − 30, has coefficients −2, −3, −14, 15, 34, −30 which has 2 changes of sign. An upper The true number of negative roots may vary from this upper bound by a bound for the number of negative roots of f(x) is two. The multiple of two (since complex number occur in conjugate pairs.) polynomial either has two negative roots or none. (As we saw in our work, x = −1 was a root twice.)

Smith (SHSU) Elementary Functions 2013 31 / 35 Smith (SHSU) Elementary Functions 2013 32 / 35 Descartes’ Rule of Signs, Negative version The Fundamental Theorem of Algebra

After the earlier material on complex numbers, we are now able to state the Fundamental Theorem of Algebra more precisely.

Descartes’ Rule of Signs narrows our search for roots of a polynomial. The Fundamental Theorem of Algebra n n−1 2 A polynomial f(x) = anx + an−1x + a2x + a1x + a0 with real Earlier we searched for roots of x3 − 8. coeﬃcients aj, has exactly n zeroes, if we include complex zeroes and also count the multiplicity of zeroes. Descartes’ Rule of Signs tells us that that polynomial has 1 positive real root and 0 negative real roots. Complex solutions come in conjugate pairs.

If we expect 3 roots then we know that the other two roots must come in Since a zero x = c of a polynomial gives a factor x − c, we can restate this complex conjugate pairs. in terms of factors.

The Fundamental Theorem of Algebra (Second version) n n−1 2 A polynomial f(x) = anx + an−1x + a2x + a1x + a0 with real coeﬃcients aj, factors completely into n linear terms, if we allow factoring involving complex numbers.

Smith (SHSU) Elementary Functions 2013 33 / 35 Smith (SHSU) Elementary Functions 2013 34 / 35 Zeroes of Polynomials

In the next lecture we explore rational functions.

(END)

Smith (SHSU) Elementary Functions 2013 35 / 35