# Sect 3.3 Zeros of Polynomials Functions

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1 101 Sect 3.3 Zeros of Polynomials Functions Objective 1: Using the Factor Theorem. Recall that a zero of a polynomial P(x) is a number c such that P(c) = 0. If the remainder of P(x) (x c) is zero, then c is a zero and x c is factor of P(x). Factor Theorem A number c is a zero of a polynomial P if and only if x c is a factor of P. Proof: ( ) Suppose c is a zero of P. Using the Division Algorithm, we can rewrite P as: P(x) = Q(x) (x c) + R(x) But, by the Remainder Theorem, R(x) = P(c) = 0 since c is a zero of P. This implies that P(x) = Q(x) (x c) which means that x c is a factor of P. ( ) Suppose x c is a factor of P. Then P(x) = Q(x) (x c) for some polynomial Q(x). If we evaluate P at x = c, we get: P(c) = Q(c) (c c) = 0 This implies that c is a zero of P. Show that the given values of c are zeros of P(x) and find all the other zeros of P(x). Ex. 1 P(x) = 2x 4 + 9x 3 9x 2 46x + 24; c = 3, 2 First, evaluate P at x = 3 and at x = 2: P( 3) = 2( 3) 4 + 9( 3) 3 9( 3) 2 46( 3) + 24 = = 0 P(2) = 2(2) 4 + 9(2) 3 9(2) 2 46(2) + 24 = = 0 Thus, 3 and 2 are zeros of P so by the factor theorem, x + 3 and x 2 are factors of P. We will now use synthetic division to break down P: Thus, P(x) = (x 2)(2x x x 12)

2 So, P(x) = (x 2)(2x x x 12) = (x 2)(x + 3)(2x 2 + 7x 4) Now, we can factor 2x 2 + 7x 4 using trial and error: 2x 2 + 7x 4 = (2x 1)(x + 4) Therefore, P(x) = (x 2)(x + 3)(2x 1)(x + 4) and the zeros of P are { 4, 3, ½, 2} Show that the given values of c are zeros of P(x) and find all the other zeros of P(x). Ex. 2 P(x) = 4x 4 19x 3 21x x + 20; c = 2, 2 First, evaluate P at x = 2 and at x = 2: P( 2) = 4( 2) 4 19( 2) 3 21( 2) ( 2) + 20 = = 0 P(2) = 4(2) 4 19(2) 3 21(2) (2) + 20 = = 0 Thus, 2 and 2 are zeros of P so by the factor theorem, x + 2 and x 2 are factors of P. We will now use synthetic division to break down P: Thus, P(x) = (x 2)(4x 3 11x 2 43x 10) So, P(x) = (x 2)(4x 3 11x 2 43x 10) = (x 2)(x + 2)(4x 2 19x 5) Now, we can factor 4x 2 19x 5 using trial and error: 4x 2 19x 5 = (4x + 1)(x 5) Therefore, P(x) = (x 2)(x + 2)(4x + 1)(x 5) and the zeros of P are { 2, ¼, 2, 5} Objective 2: Finding rational zeros of functions. In the last example, P(x) = 4x 4 19x 3 21x x + 20, the constant term was 20 and the leading coefficient was 4. The factors p of the constant term are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20 and the factors q of the leading coefficient are ± 1, ± 2, ± 4. If we form all possible ratios of the factors p of the constant term to the factors q of the leading coefficient, we get: 102

3 103 ± 1, ± 2, ± 4, ± 5, ± 10, ± 20, ± 1 2, ± 5 2, ± 1 4, and ± 5. This gives us all the 4 possible rational zeros of P(x). Notice that the zeros ¼, 2, 2, 5 are contained in this list. This leads to the following theorem. Rational Zeros Theorem Let P(x) = a n x n + a n 1 x n a 1 x + a 0 be a polynomial of degree n with integer coefficients. Then every rational zero of P is in the form p q where p q is in lowest terms and p is a factor of the constant term a 0 and q is a factor of the leading coefficient a n. Proof: Let p q be a zero of a P(x) and p q be in lowest terms. Then, P( p q ) = a n( p q )n + a n 1 ( p q )n a 1 ( p q ) + a 0 = 0 (simplify) a n p n p n 1 + a q n n a p q n 1 1 q + a 0 = 0 (multiply both sides by q n ) a n p n + a n 1 p n 1 q + + a 1 pq n 1 + a 0 q n = 0 (subtract a n p n from both sides) a n 1 p n 1 q + + a 1 pq n 1 + a 0 q n = a n p n (factor out q) q(a n 1 p n a 1 pq n 2 + a 0 q n 1 ) = a n p n This implies that q is a factor of a n p n. But since p and q are in lowest terms, then q has to be a factor of a n. We can use a similar argument to show that p is a factor of a 0. Find all the rational zeros of the P(x) and then factor P(x) into linear factors: Ex. 3 P(x) = x x3 2x x In order to use the Rational Zeros Theorem, all the coefficients have to be integers. Since the LCD of the coefficients is 6, we will first factor out 1/6 from all the terms: P(x) = x x3 2x x (Rewrite with a denominator of 6) = 6 6 x x x2 3 6 x (Factor out 1 6 ) = 1 6 (6x4 + 7x 3 12x 2 3x + 2) Now, find the possible rational zeros of 6x 4 + 7x 3 12x 2 3x + 2 The possible values of p are ± 1, ± 2, and the possible values of q are ± 1, ± 2, ± 3, ± 6. Thus, the possible combinations of p over q are

4 104 ± 1 1, ± 1 2, ± 1 3, ± 1 6, ± 2 1, ± 2 2, ± 2 3, ± 2. Simplifying and eliminating 6 duplication gives us the possible rational zeros of ± 1, ± 1 2, ± 1 3, ± 1 6, ± 2, ± 2 3. Now, we will use synthetic division to factor 6x 4 + 7x 3 12x 2 3x + 2 Let's try x = 1: Yes Thus, P(x) = 1 6 (x 1)(6x3 + 13x 2 + x 2) Let's try x = 1 again: No Now, let's try x = 1: No Let's try x = 2: No Now, let's try x = 2: Yes Hence, P(x) = 1 6 (x 1)(x + 2)(6x2 + x 1) Now, factor 6x 2 + x 1 using trial and error to get (3x 1)(2x + 1). Then, our answer is P(x) = 1 (x 1)(x + 2)(3x 1)(2x + 1). 6 This suggests a procedure for finding the rational zeros of a polynomial.

5 105 Steps to Finding the Rational Zeros 1) Use the Rational Zeros Theorem to list all possible rational zeros. 2) Use synthetic division evaluate each possible rational zero in the polynomial. If the reminder is zero, write down the quotient you found. 3) Repeat the process for the quotient you found in part 2 until you obtain a quotient that is degree two or easily factors. Factor or use the quadratic formula to find the remaining zeros. Find all the zeros of the P(x): Ex. 4 P(x) = 2x 4 x 3 67x 2 159x 63 The possible values of p are ± 1, ± 3, ± 7, ± 9, ± 21, ± 63 and the possible values of q are ± 1, ± 2. Thus, the possible combinations of p over q are: ± 1, ± 1 2, ± 3, ± 3 2, ± 7, ± 7 2, ± 9, ± 9 2 Now, we will use synthetic division Let's try x = 1: No Try x = No Try x = 3: No 21 63, ± 21, ±, ± 63, ± 2 2. Try x = Yes

6 106 Try x = 3 again Yes Now, factor 2x 2 13x 7 using trial and error: 2x 2 13x 7 = (2x + 1)(x 7). Solving (2x + 1)(x 7) = 0 yields x = ½ and x = 7. Thus, the zeros are { 3, ½, 7}. If we had to write P(x) in the last example as a product of linear factors, our answer would have been P(x) = (x + 3) 2 (2x + 1)(x 7). Notice that the factors (2x + 1) and (x 7) occur only once. In such a case, we say that the zeros ½ and 7 have multiplicity of one. Since the factor (x + 3) occurs twice, the zero 3 has multiplicity of two. Find all the real solutions to the following: Ex. 5 f(x) = x 4 + 2x 3 + 3x 2 2x 4 = 0 The possible values of p are ± 1, ± 2, ± 4 and the possible values of q are ± 1. Thus, the possible combinations of p over q are ± 1, ± 2, ± 4. Now, we will use synthetic division Let's try x = 1: Yes Try x = Yes Using the discriminant on x 2 + 2x + 4 = 0, we get: b 2 4ac = (2) 2 4(1)(4) = 4 16 = 12 which tells us that x 2 + 2x + 4 = 0 has no real solutions. Thus, the only real solutions are { 1, 1}. Each zero has multiplicity of one. Suppose in the last example, if we were asked to find all solutions (real and complex), we would have to use the quadratic formula to get the to complex solutions: x 2 + 2x + 4 = 0

7 107 x = b± b2 4ac 2a = (2) ± (2)2 4(1)(4) 2(1) = 2± 12 2 = 1 ± 3 i So, all the solutions would be { 1, 1, 1 3 i, i }. In factored form, f(x) would be (x + 1)(x 1)(x i)(x i). Notice that even though f(x) does not factor into linear factors in the real numbers, it does factor into linear factors in the complex numbers. Also, the complex zeros are what we call complex conjugates. Complex Conjugates If r = a + bi is a complex number, then r = a bi is the complex conjugate. Complex conjugate pairs have the following property: r r = (a + bi )(a bi ) = a 2 + b 2 Thus, (5 + 3i )(5 3i ) = 25 15i + 15i 9i 2 = 25 9( 1) = = 34. r and r are called conjugate pairs. Objective 3: Finding complex zeros of polynomials. The depressed equation x 2 + 2x + 4 = 0 from the last example had no real solutions. Such quadratic equations are called irreducible if they cannot be factored over the real numbers. Any irreducible quadratic equation in the real numbers can be factored in the complex numbers. This suggests that every complex non-constant polynomial has at least one complex zero. This is known as the Fundamental Theorem of Algebra. Fundamental Theorem of Algebra Every complex polynomial function with degree n 1 has at least one complex zero. Using this theorem in conjunction with the Factor Theorem leads to the following theorems: Theorem Every complex polynomial function f of degree n 1 can be factored into n linear factors (not necessarily distinct) of the form: f(x) = a n (x r 1 )(x r 2 ) (x r n ) where a n, r 1, r 2,, r n are complex numbers. In other words, every nonconstant polynomial has exact n complex zeros including repeating zeros. Conjugate Pairs Theorem Let f be a polynomial function whose coefficients are real numbers. If r = a + bi is a zero of f, then the complex conjugate r = a bi is also a zero of f so long as b 0.

8 108 Find all the complex zeros of the P(x). Then write P in factored form: Ex. 6 P(x) = x 4 + 2x x x 75 The possible values of p are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 and the possible values of q are ± 1. Thus, the possible combinations of p over q are ± 1, ± 3, ± 5, ± 15, ± 25, ± 75 which gives us the possible rational zeros. Now, we will use synthetic division: Let's try x = 1: Yes Try x = No Try x = 3: No Try x = Yes Now, factor x in the complex numbers: x = (x + 5i )(x 5i ) Solving (x + 5i )(x 5i ) = 0 yields x = ± 5i. Thus, the zeros are { 5i, 5i, 3, 1}. As a product of linear factors, we get: P(x) = (x + 5i )(x 5i )(x + 3)(x 1) Solve the following: Ex. 7 f(x) = 2x 4 + 5x 3 + 9x 2 + 2x 4 = 0 The possible values of p are ± 1, ± 2, ± 4 and the possible values of q are ± 1, ± 2. Thus, the possible combinations of p over q are:

9 109 ± 1, ± 1, ± 2, ± 4. 2 Now, we will use synthetic division Let's try x = 1: No Try x = Yes Try x = 1 again No Try x = Yes Working with the depressed equation 2x 2 + 4x + 8 = 0, we can factor out a 2 to get 2(x 2 + 2x + 4) and then we will need to use the quadratic formula on x 2 + 2x + 4: x = b± b2 4ac 2a = (2) ± (2)2 4(1)(4) 2(1) = 2± 12 2 So, the solution is { 1 3 i, i, 1, 1 2 }. If we had to write f(x) in factored form, we would get: f(x) = 2(x ( 1 3 i ))(x ( i ))(x + 1)(x 1 2 ) = 1 ± 3 i Given the zeros of a polynomial, we can construct a polynomial that would have those zeros. Let's start with the answer from example #2 and work backwards to find a polynomial.

10 110 Find a 4 th degree polynomial with the given zeros: Ex. 8 ¼, 2, 2, 5 Since ¼, 2, 2, & 5 are zeros, then (x ( ¼)) = (x + ¼), (x ( 2)) = (x + 2), (x 2), & (x 5) are factors. But saying (x + ¼) is a factor is also equivalent to saying (4x + 1) is a factor since they have the "same zero." Thus, P is the product of these factors: P(x) = (4x + 1)(x + 2)(x 2)(x 5) But by FOIL, (4x + 1)(x + 2) = 4x 2 + 9x + 2 and (x 2)(x 5) = x 2 7x + 10 Thus, P(x) = (4x + 1)(x + 2)(x 2)(x 5) = (4x 2 + 9x + 2)(x 2 7x + 10) Multiplying the two trinomials, we get: 4x 2 + 9x + 2 x 2 7x x x x 3 63x 2 14x 4x 4 + 9x 3 + 2x 2 4x 4 19x 3 21x x + 20 Hence, P(x) = 4x 4 19x 3 21x x Note that if we multiply 4x 4 19x 3 21x x + 20 by any non-zero constant, we will get another 4 th degree polynomial that has the same zeros. This means that there are an infinite number of answers for the last example; all such answers are constant multiples of one and other. Find a 4 th degree polynomial f satisfying the following conditions: Ex. 9 Zeros: 4 (multiplicity of 2), (multiplicity of 1). f( 2) = 189 (multiplicity of 1), and Since 4, 1 2, & 1 3 are zeros, then (x 4), (x ( 1 2 )) = (x ), & (x 1 3 ), are factors. But saying (x + 1 ) is a factor is also equivalent 2 to saying (2x + 1) is a factor and (x 1 ) is a factor is equivalent to 3 saying (3x 1) is a factor since they have the "same zero." Since the zero of 4 has a multiplicity of 2, the factor (x 4) will occur twice. Thus, f is the product of these factors: f(x) = (x 4) 2 (2x + 1)(3x 1)

11 But by FOIL, (x 4)(x 4) = x 2 8x + 16 and (2x + 1)(3x 1) = 6x 2 + x 1 Thus, f(x) = (x 4) 2 (2x + 1)(3x 1) = (x 2 8x + 16)(6x 2 + x 1) Multiplying the two trinomials, we get: x 2 8x x 2 + x 1 x 2 + 8x 16 x 3 8x x 6x 4 48x x 2 6x 4 47x x x 16 So, f(x) = a(6x 4 47x x x 16). But, f( 2) = 189, which says: f( 2) = a(6( 2) 4 47( 2) ( 2) ( 2) 16) = 189 a( ) = a = 189 or a = ¼ Hence, f(x) = ¼(6x 4 47x x x 16) = 3 2 x x x2 6x Find a polynomial function with the given information: Ex. 10 Degree 5; Zeros: 2, 3 5i, 4i Since the polynomial has a degree of 5, then it has five zeros. The conjugate pair of 4i is 4i which is also a zero and the conjugate pair of 3 5i is 3 + 5i which is a zero as well. So, the missing zeros are 4i and 3 + 5i. This means that (x ( 2)), (x (3 5i )), (x (3 + 5i )), (x ( 4i )), & (x (4i )) are factors. Thus, f can be written as: f(x) = a(x ( 2))(x (3 5i ))(x (3 + 5i ))(x ( 4i ))(x (4i )) where a is any nonzero real number. f(x) = a(x + 2)[x (3 5i )][x (3 + 5i )](x + 4i )(x 4i ) (expand) f(x) = a(x + 2)[x 2 (3 5i i )x + (3 5i )(3 + 5i )](x 2 16i 2 ) f(x) = a(x + 2)[x 2 6x i 2 ](x 2 16i 2 ) (replace i 2 by 1) f(x) = a(x + 2)[x 2 6x ](x ) f(x) = a(x + 2)(x 2 6x + 34)(x ) f(x) = a(x 3 4x x + 68)(x ) f(x) = a(x 5 4x x 3 + 4x x )

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