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**Real Zeros of Polynomial Functions Real Zeros of Polynomial Functions**

Polynomial Zeros Real Zeros of Polynomial Functions This section explores some of the techniques employed to locate and evaluate the real zeros of polynomial functions. Discussion of complex zeros is left for later sessions on the Fundamental Theorem of Algebra. This module should take at least four class sessions. Additional examples should be worked out with class participation. The step from concrete examples to the more abstract general case is often difficult for inexperienced student. Having them write down the general case steps and justify each can help to promote understanding of the basic concepts. Working out concrete examples enhances that understanding. Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Even/Odd Multiplicity Examples x y(x) x y(x) x y(x) y = x + 3 ● ● ● ● y = (x – 3)2 y = (x – 3)3 y = (x – 3)4 x y(x) x y(x) x y(x) Even/Odd Multiplicities Multiplicities of zeros for polynomials of even or odd degree give the graphs different behaviors in a small neighborhood of the zero. In the examples shown from left to right, the zeros of the polynomials have multiplicities of 1, 2, 3, 4, and 5. The last two polynomials each have two distinct zeros, -3 and +3, with multiplicity 3 and 1, respectively, and -2 and +3 with multiplicities 3 and 2, respectively. We can use the multiplicity of each zero to tell something about the behavior of the graph near the zero. In the cases of odd multiplicity (zeros 1, 3, 5, -2 and -3), as we move away from the zero in both positive and negative directions, the graph rises on one side of the zero and falls on the other. In other words, in any small interval containing the zero, the graph is either increasing or decreasing throughout the interval. In the cases of even multiplicity (zero at x = 3, multiplicities 2 and 4), as we move away from the zero in both directions, the graph either rises on both sides or falls on both sides. In other words, the zero is a turning point for the graph, so that it changes from increasing to decreasing or from decreasing to increasing at the zero. These facts are very helpful in sketching the graphs of polynomials to give a general impression of its overall behavior. This is quite often all that is needed in understanding the behavior of the function. ● ● ● ● ● y = (x – 3)5 y = (x + 3)3(x – 3) y = (x + 2)3(x – 3)2 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test Let f(x) = anxn + … + a2x2 + a1x + a0 for integer coefficients with an ≠ 0 Then all rational zeros of f(x) are of form p/q p is a factor of a0 and q is a factor of an p and q have no common factors The Rational Zero Test This illustrates a means for narrowing the search for rational zeros. It does not apply to irrational zeros and it does apply only to polynomial functions with integer coefficients. This is sometimes called the Rational Root Test. It is important to understand the limitations of the technique, since application of the methodology to other types of functions yields unpredictable results and is one of the most common errors in the application of mathematics – trying to apply principles where the do not apply. The reason that rational zeros are desirable is that they are exact values. If the zeros are not rational numbers, then other methods must be applied to find approximations for the zeros, which can lead to larger errors in the final outcome of larger problems. In many cases, some of the zeros are rational and some are irrational. If we can find all of the rational zeros with a simple method, then the search for the irrational zeros is a much smaller problem. The whole purpose in using this method is to reduce the number of operations needed to find the zeros of a polynomial. The fact that the form of a rational number is p/q where p is a factor (or divisor) of a0 the constant term, and q is a factor (or divisor) of an the lead coefficient, allows us to rule out many other possible values for the zeros. Note that if we divide f(x) by an the constant term in the new polynomial is a0/an compared with rational zeros p/q where p divides a0 and q divides an. So p/q is a factor of a0/an. We will find that the fact that equal polynomials have the same factors very useful in finding the rational zeros of a polynomial. 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test f(x) = anxn + … + a2x2 + a1x + a0 All rational zeros of f(x) of form p/q , with p is a factor of a0 and q is a factor of an NOTE: This works only for integer coefficients NOT all zeros are rational numbers NO irrational zeros of f(x) are included The Rational Zero Test This illustrates a means for narrowing the search for rational zeros. It does not apply to irrational zeros and it does apply only to polynomial functions with integer coefficients. This is sometimes called the Rational Root Test. It is important to understand the limitations of the technique, since application of the methodology to other types of functions yields unpredictable results and is one of the most common errors in the application of mathematics – trying to apply principles where the do not apply. The reason that rational zeros are desirable is that they are exact values. If the zeros are not rational numbers, then other methods must be applied to find approximations for the zeros, which can lead to larger errors in the final outcome of larger problems. In many cases, some of the zeros are rational and some are irrational. If we can find all of the rational zeros with a simple method, then the search for the irrational zeros is a much smaller problem. The whole purpose in using this method is to reduce the number of operations needed to find the zeros of a polynomial. The fact that the form of a rational number is p/q where p is a factor (or divisor) of a0 the constant term, and q is a factor (or divisor) of an the lead coefficient, allows us to rule out many other possible values for the zeros. Note that if we divide f(x) by an the constant term in the new polynomial is a0/an compared with rational zeros p/q where p divides a0 and q divides an. So p/q is a factor of a0/an. We will find that the fact that equal polynomials have the same factors very useful in finding the rational zeros of a polynomial. 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Zeros and Factors FACT: If two polynomials are equal then they have the same factors If f(x) = (x – k1)Q1(x) and if Q1(x) = (x – k2)Q2(x) then we have f(x) = (x – k1)(x – k2)Q2(x) The Rational Zero Test This illustrates a means for narrowing the search for rational zeros. It does not apply to irrational zeros and it does apply only to polynomial functions with integer coefficients. This is sometimes called the Rational Root Test. It is important to understand the limitations of the technique, since application of the methodology to other types of functions yields unpredictable results and is one of the most common errors in the application of mathematics – trying to apply principles where the do not apply. The reason that rational zeros are desirable is that they are exact values. If the zeros are not rational numbers, then other methods must be applied to find approximations for the zeros, which can lead to larger errors in the final outcome of larger problems. In many cases, some of the zeros are rational and some are irrational. If we can find all of the rational zeros with a simple method, then the search for the irrational zeros is a much smaller problem. The whole purpose in using this method is to reduce the number of operations needed to find the zeros of a polynomial. The fact that the form of a rational number is p/q where p is a factor (or divisor) of a0 the constant term, and q is a factor (or divisor) of an the lead coefficient, allows us to rule out many other possible values for the zeros. Note that if we divide f(x) by an the constant term in the new polynomial is a0/an compared with rational zeros p/q where p divides a0 and q divides an. So p/q is a factor of a0/an. We will find that the fact that equal polynomials have the same factors very useful in finding the rational zeros of a polynomial. 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test Example Factor completely: f(x) = 3x4 – 12x3 – 24x2 + 36x + 45 = 3(x4 – 4x3 – 8x2 + 12x + 15) = 3g(x) Here an = a4 = 1 and a0 = 15 Factors p of 15 are: ±1, ±3, ±5, ±15 ; Factors q of 1 are: ±1 Possible p/q values are: ±1, ±3, ±5, ±15 The Rational Zero Test Example This example includes both rational and irrational zeros. We first check to see that f(x) has integer coefficients, since the Rational Zeros Theorem applies only in this case. We next factor out the lead coefficient 3, leaving a simpler polynomial factor of g(x) to analyze. We then proceed with finding the factors of a0 = 15 and a4 = 1. These are 1 , 3 , and 5, for a0 and 1 for a4 = 1. The possible rational zeros of g(x) are then 1 , 3 , and 5. We test each of these in turn with synthetic division. Note that k = 1 and k = 3 are not zeros (non-zero remainder) so that (x – 1) and (x – 3) are NOT factors of g(x). However, k = 5 IS a zero of g(x), which, using the Factor Theorem, allows us to write g(x) = (x – 5)Q1(x) where Q1(x) is the quotient resulting from dividing g(x) by (x – 5). Since any factor of Q1(x) will also be a factor of g(x), we try our next possible rational zero, namely x = –1, on Q1(x). The zero remainder shows that k = –1 is a zero for Q1(x) and hence for g(x), leaving a quotient Q2(x) = x2 – 3. We can now write g(x) = (x – 5)(x + 1)Q2(x) = (x – 5)(x + 1)(x2 – 3) which we continue to analyze. Using the Rational Zero Test quickly gains us two zeros for f(x), thereby narrowing the search for the remaining zeros. 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test Example g(x) = x4 – 4x3 – 8x2 + 12x + 15 Possible p/q values: ±1, ±3, ±5, ±15 Check zeros of g(x) k = 1 1 1 –4 – 1 –3 –11 1 (x – 1) is NOT a factor of g(x) 1 –3 –11 1 16 The Rational Zero Test Example This example includes both rational and irrational zeros. We first check to see that f(x) has integer coefficients, since the Rational Zeros Theorem applies only in this case. We next factor out the lead coefficient 3, leaving a simpler polynomial factor of g(x) to analyze. We then proceed with finding the factors of a0 = 15 and a4 = 1. These are 1 , 3 , and 5, for a0 and 1 for a4 = 1. The possible rational zeros of g(x) are then 1 , 3 , and 5. We test each of these in turn with synthetic division. Note that k = 1 and k = 3 are not zeros (non-zero remainder) so that (x – 1) and (x – 3) are NOT factors of g(x). However, k = 5 IS a zero of g(x), which, using the Factor Theorem, allows us to write g(x) = (x – 5)Q1(x) where Q1(x) is the quotient resulting from dividing g(x) by (x – 5). Since any factor of Q1(x) will also be a factor of g(x), we try our next possible rational zero, namely x = –1, on Q1(x). The zero remainder shows that k = –1 is a zero for Q1(x) and hence for g(x), leaving a quotient Q2(x) = x2 – 3. We can now write g(x) = (x – 5)(x + 1)Q2(x) = (x – 5)(x + 1)(x2 – 3) which we continue to analyze. Using the Rational Zero Test quickly gains us two zeros for f(x), thereby narrowing the search for the remaining zeros. k = 3 3 1 –4 – 3 –3 –33 –63 (x – 3) is NOT a factor of g(x) 1 –1 –11 –21 –48 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test Example Possible p/q values: ±1, ±3, ±5, ±15 k = 5 5 1 –4 – 5 5 –15 –15 (x – 5) IS a factor of g(x) 1 1 –3 –3 Q1(x) The Rational Zero Test Example This example includes both rational and irrational zeros. We first check to see that f(x) has integer coefficients, since the Rational Zeros Theorem applies only in this case. We next factor out the lead coefficient 3, leaving a simpler polynomial factor of g(x) to analyze. We then proceed with finding the factors of a0 = 15 and a4 = 1. These are 1 , 3 , and 5, for a0 and 1 for a4 = 1. The possible rational zeros of g(x) are then 1 , 3 , and 5. We test each of these in turn with synthetic division. Note that k = 1 and k = 3 are not zeros (non-zero remainder) so that (x – 1) and (x – 3) are NOT factors of g(x). However, k = 5 IS a zero of g(x), which, using the Factor Theorem, allows us to write g(x) = (x – 5)Q1(x) where Q1(x) is the quotient resulting from dividing g(x) by (x – 5). Since any factor of Q1(x) will also be a factor of g(x), we try our next possible rational zero, namely x = –1, on Q1(x). The zero remainder shows that k = –1 is a zero for Q1(x) and hence for g(x), leaving a quotient Q2(x) = x2 – 3. We can now write g(x) = (x – 5)(x + 1)Q2(x) = (x – 5)(x + 1)(x2 – 3) which we continue to analyze. Using the Rational Zero Test quickly gains us two zeros for f(x), thereby narrowing the search for the remaining zeros. k = –1 –1 –3 –3 –1 3 (x + 1) IS a factor of Q1(x) 1 –3 Q2(x) 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test Example Thus far f(x) = 3g(x) = 3(x – 5)Q1(x) = 3(x – 5)(x + 1)Q2(x) = 3(x – 5)(x + 1)(x2 – 3) Q2(x) = x2 – 3 with possible rational zeros of ±1 and ±3 The Rational Zero Test Example This example includes both rational and irrational zeros. We first check to see that f(x) has integer coefficients, since the Rational Zeros Theorem applies only in this case. We next factor out the lead coefficient 3, leaving a simpler polynomial factor of g(x) to analyze. We then proceed with finding the factors of a0 = 15 and a4 = 1. These are 1 , 3 , and 5, for a0 and 1 for a4 = 1. The possible rational zeros of g(x) are then 1 , 3 , and 5. We test each of these in turn with synthetic division. Note that k = 1 and k = 3 are not zeros (non-zero remainder) so that (x – 1) and (x – 3) are NOT factors of g(x). However, k = 5 IS a zero of g(x), which, using the Factor Theorem, allows us to write g(x) = (x – 5)Q1(x) where Q1(x) is the quotient resulting from dividing g(x) by (x – 5). Since any factor of Q1(x) will also be a factor of g(x), we try our next possible rational zero, namely x = –1, on Q1(x). The zero remainder shows that k = –1 is a zero for Q1(x) and hence for g(x), leaving a quotient Q2(x) = x2 – 3. We can now write g(x) = (x – 5)(x + 1)Q2(x) = (x – 5)(x + 1)(x2 – 3) which we continue to analyze. Using the Rational Zero Test quickly gains us two zeros for f(x), thereby narrowing the search for the remaining zeros. Synthetic division shows none of these are zeros of Q2(x) 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Rational Zero Test Example Q2(x) = x2 – 3 = 3 x2 – ( ) 2 = 3 (x – ) (x ) Thus f(x) = 3(x – 5)(x + 1)Q2(x) 3 (x – ) (x ) = 3(x – 5)(x + 1) The Rational Zero Test Example This example includes both rational and irrational zeros. We first check to see that f(x) has integer coefficients, since the Rational Zeros Theorem applies only in this case. We next factor out the lead coefficient 3, leaving a simpler polynomial factor of g(x) to analyze. We then proceed with finding the factors of a0 = 15 and a4 = 1. These are 1 , 3 , and 5, for a0 and 1 for a4 = 1. The possible rational zeros of g(x) are then 1 , 3 , and 5. We test each of these in turn with synthetic division. Note that k = 1 and k = 3 are not zeros (non-zero remainder) so that (x – 1) and (x – 3) are NOT factors of g(x). However, k = 5 IS a zero of g(x), which, using the Factor Theorem, allows us to write g(x) = (x – 5)Q1(x) where Q1(x) is the quotient resulting from dividing g(x) by (x – 5). Since any factor of Q1(x) will also be a factor of g(x), we try our next possible rational zero, namely x = –1, on Q1(x). The zero remainder shows that k = –1 is a zero for Q1(x) and hence for g(x), leaving a quotient Q2(x) = x2 – 3. We can now write g(x) = (x – 5)(x + 1)Q2(x) = (x – 5)(x + 1)(x2 – 3) which we continue to analyze. Using the Rational Zero Test quickly gains us two zeros for f(x), thereby narrowing the search for the remaining zeros. Two rational zeros and two irrational zeros 3 x = 5, –1, ± 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Equations New functions f(x) lead to new types of equations to solve Set f(x) = 0 and find the zeros Examples Find all real solutions of: 1. x4 – 1 = 0 Equations Each new function f(x) we define leads to another kind of equation to solve when we set f(x) = 0 and find the zeros of the function. Here are some example problems to check one’s grasp of the methods just presented. These can be used merely as examples to work out in class or as exercises for the student as homework. In Example 1 we see the difference of fourth powers actually solvable using the difference of squares. One way of seeing this is to use the substitution u = x2 and rewrite as u2 – 1 = 0. This yields u = 1 and u = –1, that is , x2 = 1 and x2 = –1. The first of these yields y = 1 and y = -1. The second yields no real solutions, though there are two imaginary solutions. These will be studied later. In Example 2, we rewrite as x3 – x = 0 and factor: x(x2 – 1) = x(x – 1)(x + 1). So, x = 0, x = 1 and x = -1. In Example 3, we again make the substitution u = x2 and rewrite: u2 – 5u + 4 = 0 or (u – 4)(u – 1) = 0 which yields x2 – 4 = 0 and x2 – 1 = 0, giving x = 1, x = –1, x = 2, x = –2 . In Example 4, we again make a substitution: u = x3 so that we have u2 – 19u – 216 = 0 giving (u – 27)(u + 8) = (x3 – 33)(x3 + 23) = 0. There are only two real solutions: x = -2 and x = 3. The other four solutions are complex numbers. 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Equations Examples Find all real solutions of: 2. x3 = x 3. x4 – 5x2 + 4 = 0 4. x6 – 19x3 = 216 Equations Each new function f(x) we define leads to another kind of equation to solve when we set f(x) = 0 and find the zeros of the function. Here are some example problems to check one’s grasp of the methods just presented. These can be used merely as examples to work out in class or as exercises for the student as homework. In Example 1 we see the difference of fourth powers actually solvable using the difference of squares. One way of seeing this is to use the substitution u = x2 and rewrite as u2 – 1 = 0. This yields u = 1 and u = –1, that is , x2 = 1 and x2 = –1. The first of these yields y = 1 and y = -1. The second yields no real solutions, though there are two imaginary solutions. These will be studied later. In Example 2, we rewrite as x3 – x = 0 and factor: x(x2 – 1) = x(x – 1)(x + 1). So, x = 0, x = 1 and x = -1. In Example 3, we again make the substitution u = x2 and rewrite: u2 – 5u + 4 = 0 or (u – 4)(u – 1) = 0 which yields x2 – 4 = 0 and x2 – 1 = 0, giving x = 1, x = –1, x = 2, x = –2 . In Example 4, we again make a substitution: u = x3 so that we have u2 – 19u – 216 = 0 giving (u – 27)(u + 8) = (x3 – 33)(x3 + 23) = 0. There are only two real solutions: x = -2 and x = 3. The other four solutions are complex numbers. 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Problem Solving Find the width W of the rectangle from its length and area A. Also determine W when L = 5 inches. A = WL = W(x2 + 1) = 3x3 + 3x – 5x2 – 5 = 3x(x2 + 1) – 5(x2 + 1) = (3x – 5)(x2 + 1) W A = 3x3 – 5x2 + 3x – 5 Problem Solving Here is a nice little problem using polynomials. It can be solved by factoring A, as shown in the illustration, or by directly attacking the division of A by x2 + 1 using conventional long division. Question: Could we use synthetic division instead of long division? YES! As shown, long division is not necessary, since the polynomial A representing the area can be factored and one of the factors is exactly the divisor L = x Thus simply isolating x2 + 1 in both numerator and denominator allows us to replace x2 + 1 divided by itself with 1. As shown, this leaves W = 3x – 5. Clearly then if we let x = 5 we get W = 10. L = x2 + 1 A 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Problem Solving A = (3x – 5)(x2 + 1) W A = 3x3 – 5x2 + 3x – 5 L = x2 + 1 W = x2 + 1 A = x2 + 1 (3x – 5)(x2 + 1) = 3x – 5 At x = 5 inches, Problem Solving Here is a nice little problem using polynomials. It can be solved by factoring A, as shown in the illustration, or by directly attacking the division of A by x2 + 1 using conventional long division. Question: Could we use synthetic division instead of long division? YES! As shown, long division is not necessary, since the polynomial A representing the area can be factored and one of the factors is exactly the divisor L = x Thus simply isolating x2 + 1 in both numerator and denominator allows us to replace x2 + 1 divided by itself with 1. As shown, this leaves W = 3x – 5. Clearly then if we let x = 5 we get W = 10. W = 3(5) – 5 = 10 inches Note: To find W we could have used conventional long division 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Think about it ! 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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**Real Zeros of Polynomial Functions**

Note: To find W we could have used conventional long division W = L A 3x3 – 5x2 + 3x – 5 x2 + 1 = x2 + 1 3x3 – 5x2 + 3x – 5 Problem Solving Here is a nice little problem using polynomials. It can be solved by factoring A, as shown in the illustration, or by directly attacking the division of A by x2 + 1 using conventional long division. Question: Could we use synthetic division instead of long division? YES! As shown, long division is not necessary, since the polynomial A representing the area can be factored and one of the factors is exactly the divisor L = x Thus simply isolating x2 + 1 in both numerator and denominator allows us to replace x2 + 1 divided by itself with 1. As shown, this leaves W = 3x – 5. Clearly then if we let x = 5 we get W = 10. x2 + 1 L = x2 + 1 7/23/2013 Polynomial Zeros Polynomial Zeros 7/23/2013

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