Chapter 3 Section 3.3 Real Zeros of Polynomial Functions

10/29/2010 Section 3.3 v5.0 2 Special Forms Difference of Squares (Conjugate Binomials) a 2 – b 2 = ( a – b )( a + b ) Difference of n th Powers a n – b n = ( a – b )( a n–1 + a n–2 b + a n–3 b ab n–1 + b n–1 ) Perfect-Square Trinomials ( a + b ) 2 = a ab + b 2 ( a – b ) 2 = a 2 – 2 ab + b 2 Binomial Theorem Polynomial Functions n(n – 1) 2 a n–2 b bnbn n(n–1)...(2) (n – 1)! ab n–1 ( a + b ) n = a n + n a n–1 b +

10/29/2010 Section 3.3 v5.0 3 Binomial Theorem Expansion of ( a + b ) n Examples: ( a + b ) 2 = a ab + b 2 ( a + b ) 3 = a a 2 b + 3 ab 2 + b 3 ( a + b ) 4 = a a 3 b + 6 a 2 b ab 3 + b 4 ( a + b ) 5 = 1∙ a 5 + a 4 b + a 3 b 2 + a 2 b 3 + ab 4 + b 5 = a a 4 b + 10 a 3 b a 2 b ab 4 + b 5 n(n – 1) 2 ( a + b ) n = a n + n a n–1 b + a n–2 b 2 + … + bnbn = ∑ k=0 n (n–k)! n! k! a n–k b k 5∙4 2∙1 5∙4∙3 3∙2∙1 5∙4∙3∙2 4∙3∙2∙1 5! 5

10/29/2010 Section 3.3 v5.0 4 Expansion of ( a + b ) n Pascal’s Triangle Coefficients for ( a + b ) n Examples: ( a + b ) 2 = 1∙ a ab + 1∙ b 2 ( a + b ) 3 = 1∙ a a 2 b + 3 ab 2 + 1∙ b 3 ( a + b ) 4 = 1∙ a a 3 b + 6 a 2 b ab 3 + 1∙ b 4 ( a + b ) 6 = Binomial Theorem n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 1∙ a a 5 b + 15 a 4 b a 3 b a 2 b ab 5 + 1∙ b Fibonacci Sequence Degree

10/29/2010 Section 3.3 v5.0 5 Division of Polynomials Long Division (by hand): Can we do this with polynomials ? Writing the divisor and dividend as a base-10 polynomial we do the long division again Polynomial Functions 245 ) Quotient Remainder = We just did ! So = (10 2 ) + 4(10 1 ) + 5 9(10 3 ) + 8(10 2 ) + 1(10 1 ) + 5 ) 4(10 1 ) 8(10 3 ) + 16(10 2 ) + 20(10 1 ) 9(10 3 ) + 8(10 2 ) + 0(10 1 ) 1(10 1 ) Adjust powers 8(10 3 ) + 18(10 2 ) + 0(10 1 ) Quotient Remainder

10/29/2010 Section 3.3 v5.0 6 Division of Polynomials In general for polynomial functions f(x) and d(x), d(x) ≠ 0 where Q(x) is the quotient, r(x) is the remainder, d(x) is the divisor and f(x) is the dividend Thus where either r(x) = 0 or deg r(x) < deg d(x) Polynomial Functions + f(x) d(x) = Q(x) r(x) d(x) f(x) = d(x) ∙ Q(x) + r(x)

10/29/2010 Section 3.3 v5.0 7 Division by Monomials Example Polynomial Functions 12x x 3 – 9x 2 + 6x – 2 3x 2 = 12x 4 3x x 3 3x 2 9x 2 3x 2 – 6x 3x – = 4x 2 + 9x – 3 2 3x 2 – 2 x + = 4x 2 + 9x – 3 3x 2 2 – 6x + Quotient Remainder OR 12x x 3 – 9x 2 + 6x – 2 3x 2 ) 4x 2 12x 4 27x 3 + 9x 27x 3 – 9x 2 – 3 – 9x 2 6x – 2 Remainder Quotient Note: deg r(x) = 1 < deg d(x) = 2

10/29/2010 Section 3.3 v5.0 8 Division by Linear Binomials Example Polynomial Functions x 4 + 3x 3 – 4x + 1 x + 2 ) x3x3 x 4 + 2x 3 x3x3 + x 2 x 3 + 2x 2 – 2x 2 – 2x – 2x 2 – 4x 1 – 4x Quotient Remainder Thus x 4 + 3x 3 – 4x + 1 x = x 3 + x 2 – 2x x Question: Can we do this faster or more simply ? Note: deg r(x) = 0 < deg d(x) = 1

10/29/2010 Section 3.3 v5.0 9 Synthetic Division Example The arithmetic operations involve only the coefficients So, using synthetic division we deal only with the coefficients Polynomial Functions 2x 4 – 3x 3 + 5x 2 + 4x + 3 x + 2 ) 2x 3 2x 4 + 4x 3 –7x 3 – 7x 2 –7x 3 – 14x 2 19x 2 – 34 19x x – 34x + 4x +19x + 5x – 34x – – –7 – –34 –68 71 Subtract 2 – –2 2 –4 – –38 – Add d(x) = x + k = x + 2 d(x) = x – k = x – (–2) NOTE: deg r(x) = 0 < 1 = deg d(x) Remainder

10/29/2010 Section 3.3 v Degree Facts For any polynomials A(x), B(x) deg ( A(x)  B(x) ) deg ( A(x) + B(x) ) Examples: deg ( (x 2 + 1)  (3x 3 – 4x 2 + 5x + 7) ) deg ( (x – 1)  (–2x 5 + x – 5) ) deg ( (3x 2 – 4) + (2x 4 + 6x + 3) ) deg ( (4x 3 – 7x 2 – 2x +11) + (3x 3 + x – 10) ) Polynomial Functions = deg A(x) + deg B(x) = max { deg A(x), deg B(x) } = = 5 = = 6 = max { 2, 4 } = 4 = max { 3, 3 } = 3

10/29/2010 Section 3.3 v Division Algorithm for Polynomials If f(x) and d(x) are polynomial functions with 0 < deg d(x) < deg f(x) then there exist unique polynomial functions Q(x) and r(x) such that f(x) = d(x) ∙ Q(x) + r(x) where either r(x) = 0 or deg r(x) < deg d(x) Polynomial Functions Note: This just says that f(x) d(x) Q(x) =+ r(x) d(x) Example x 2 – 1 x – 1 = x + 1 deg f(x) = 2, deg d(x) = 1, r(x) = 0 Thus f(x) = d(x) ∙ Q(x) + r(x) becomes x 2 – 1 = (x – 1) ∙ (x + 1) + 0

10/29/2010 Section 3.3 v Division Algorithm and Degrees Given f(x) = d(x) ∙ Q(x) + r(x) where either r(x) = 0 or deg r(x) < deg d(x) Question: If deg d(x) = m > 0, deg Q(x) = n, and deg f(x) = p > m, what is the relationship among m, n and p ? Note: deg ( d(x)  Q(x) ) = deg d(x) + deg Q(x) = m + n deg r(x) < deg d(x) = m deg ( d(x)  Q(x) + r(x) ) = deg ( d(x)  Q(x) ) = m + n p = deg f(x) = deg ( d(x)  Q(x) + r(x) ) = m + n Thus p = m + n Polynomial Functions... could have deg r(x) = 0

10/29/2010 Section 3.3 v Remainders and Functional Values In case the divisor d(x) is linear, i.e. d(x) = x – k, then becomes where deg r(x) < deg (x – k) i.e. r(x) is just a constant r At x = k this becomes So the value of f at x = k, or f(k), is just the remainder when f(x) is divided by x – k Polynomial Functions f(x) = d(x)  Q(x) + r(x) f(x) = (x – k)Q(x) + r(x) f(k) = (k – k)Q(k) + r f(x) = (x – k)Q(x) + r... which forces deg r(x) = 0... and thus … and thus … f(k) = r = 1 = r

10/29/2010 Section 3.3 v Remainders and Functional Values (continued) Example: By synthetic division we found that dividing yields Thus f(–2) = 71 To evaluate this by the usual “brute force” method f(–2) = 2(–2) 4 – 3(–2) 3 + 5(–2) 2 + 4(–2) + 3 = 2(16) – 3(–8) + 5(4) – = – = 71 Polynomial Functions d(x) = (x + 2) = (x – (–2)) f(x) = 2x 4 – 3x 3 + 5x 2 + 4x + 3 by + 71 Question: Which method is easier? This leads to... f(x) = (2x 3 – 7x 2 – 34)+ 19x =(x + 2) Q(x)(x + 2) + r

10/29/2010 Section 3.3 v The Remainder Theorem Polynomial f(x) divided by x – k yields a remainder of f(k) Question: What if the remainder is 0 ? Then we know that f(x) = (x – k)Q(x) + r(x) and thus (x – k) is a factor of f(x) The Factor Theorem Polynomial f(x) has factor x – k if and only if f(k) = 0 Polynomial Functions = (x – k)Q(x) … leading to …

10/29/2010 Section 3.3 v Factor Theorem Corollary If (x – k) is a factor of f(x) = a n x n + a n–1 x n – a 1 x + a 0 then k is a factor of a 0 Note that f(x) = (x – k)Q 1 (x) Polynomial Functions Question: What about the converse ? WHY ? = xQ 1 (x) – kQ 1 (x) = xQ 1 (x) – k(b n–1 x n–1 + … + b 1 x + b 0 ) = [ a n x n + a n–1 x n–1 + … + a 1 x ] + a 0 = [ xQ 1 (x) – k(b n–1 x n–1 + b 1 x) ] – kb 0 Since two polynomials are equal if and only if corresponding terms are equal, then a 0 = – kb 0 Clearly, then, k is a factor of a 0

10/29/2010 Section 3.3 v Factor Theorem Corollary (continued) Is the converse true? Example: f(x) = x 4 + x 3 – 19x x + 30 Polynomial Functions 1 1 – –13 –26 –15 –30 0 Thus (x – 2) is a factor of f(x) Try k = 6... and 2 is a factor of – So (x – 6) is NOT a factor of f(x)... BUT 6 IS a factor of 30 Try k = 2 Question: What if k = ±1, –2, ±3, ±5, –6, ±10, ±15 ? What about other factors of 30 ?

10/29/2010 Section 3.3 v Factor Theorem Example Given the graph of polynomial f(x) Estimate the degree of f(x) Even or odd degree ? Find the factors of f(x) Since f(–7) = f(2) = f(11) = 0 then factors are (x + 7), (x – 2), (x – 11) Note: (x + 7)(x – 2)(x – 11) = x 3 – 6x 2 – 69x Question: Polynomial Functions x y ● ● ● (–7, 0) (2, 0) (11, 0) Is f(x) = (x + 7)(x – 2)(x – 11) ? Not necessarily ! Note: f(x) = (x + 7)Q 1 (x) What is Q 3 (x) ? Is Q 3 (x) constant ? Does Q 3 (x) have factors x – k ? = (x + 7)(x – 2)Q 2 (x)= (x + 7)(x – 2)(x – 1)Q 3 (x)

10/29/2010 Section 3.3 v Factor Theorem Example Given the graph of polynomial f(x) estimate the degree of f(x) Even or odd degree ? Find the factors of f(x) f(–3) = f(5) = 0 so (x + 3) and (x – 5) are factors Note: (x + 3)(x – 5) = x 2 – 2x 2 – 15 Question: Polynomial Functions x y (–3, 0) (5, 0) ● ● Is f(x) equal to x 2 – 2x – 15 ? Probably not ! Note: f(x) = (x + 3)Q 1 (x) = (x + 3)(x – 5)Q 2 (x) Probably, but which ones and how many ? Does Q 2 (x) have factors (x – k) ? What does the graph of x 2 – 2x – 15 look like ?

10/29/2010 Section 3.3 v Intercepts, Zeros and Factors These things are related If f(k) = 0, then Point (k, 0) on the graph of f is an x-intercept The number k is a zero for f(x), i.e. f(k) = 0 (x – k) is a factor of f(x) The number k is a factor of the constant term of f(x) Polynomial Functions WHY ?

10/29/2010 Section 3.3 v Completely Factored Polynomials Example: Factor f(x) = 2x x x 2 – 54x – 108 Now use synthetic division to check out zeros Polynomial Functions = 2(x 4 + 7x 3 + 9x 2 – 27x – 54) 1 –3 4 –12 9 – –27 –54 –3 x – k = x – (–3) (x + 3) is a factor 1 – –3 –18 –3 –6 (x + 3) is a factor 1 –3 – –6 –3 (x + 3) is a factor (x – 2) is a factor f(x) = 2(x + 3) 3 (x – 2) Note: x = –3 is a repeated zero of multiplicity 3 Q1Q1 Q2Q2

10/29/2010 Section 3.3 v Complete Factoring with Multiple Zeros General Form f(x) = a n x n + a n–1 x n– a 1 x + a 0 = a n (x – k n )(x – k n–1 )... (x – k 1 ) where some of the zeros k i may be repeated Degree of f(x) = n Number of real zeros m, m ≤ n If all zeros occur at x-intercepts then, counting multiplicities, total number of zeros is n Example f(x) = 2(x + 3) 3 (x – 2) one zero at 2 and one at –3 of multiplicity 3 total zeros, counting multiplicities, = 4 = deg f(x) Polynomial Functions, i.e. are real numbers,

10/29/2010 Section 3.3 v Complete Factoring Examples 1. f(x) = 5x 3 – 10x 2 – 15x = 5x(x 2 – 2x – 3) = 5x(x – 3)(x + 1) = 5(x + 1)(x – 0)(x – 3) 2. Given: f(x) is a quadratic polynomial lead coefficient is 7 f(–3) = 0 and f(2) = 0 Write f(x) in complete factored form Note that –3 and 2 are zeros of f(x) From the Factor Theorem x –(–3) and x – 2 are factors of f(x) Thus f(x) = 7(x + 3)(x – 2) Polynomial Functions

10/29/2010 Section 3.3 v Even/Odd Multiplicities Examples Polynomial Functions x y(x) x x x x ● y = (x – 3) 2 ● y = x + 3 ● y = (x – 3) 3 ● y = (x – 3) 4 ● y = (x – 3) 5 ● ● y = (x + 3) 3 (x – 3) x y(x) y = (x + 2) 3 (x – 3) 2 ● ●

10/29/2010 Section 3.3 v Rational Zero Test Let f(x) = a n x n + … + a 2 x 2 + a 1 x + a 0 where a n ≠ 0 and all coefficients are integers Then all rational zeros of f(x) are of form p/q, in lowest terms, where p is a factor of a 0 and q is a factor of a n NOTE: This works only if the coefficients are integers It does NOT say all zeros are rational numbers It does NOT include any irrational zeros of f(x) FACT: If two polynomials are equal they have the same factors If f(x) = (x – k 1 )Q 1 (x) and if Q 1 (x) = (x – k 2 )Q 2 (x) then we have f(x) = (x – k 1 )(x – k 2 )Q 2 (x) Polynomial Functions

10/29/2010 Section 3.3 v k = –1 Rational Zero Test Example Factor completely: f(x) = 3x 4 – 12x 3 – 24x x + 45 f(x) = 3(x 4 – 4x 3 – 8x x + 15) = 3g(x) Here a n = a 4 = 1 and a 0 = 15 Factors p of 15 are:  1,  3,  5,  15 ; Factors q of 1 are:  1 Possible rational zeros p/q are:  1,  3,  5,  15 Check zeros of g(x) with synthetic division: Polynomial Functions 1 –1 1 –4 – –3 – k = –4 – –1 –3 –11 –33 –21 k = 3 –63 – –4 – –3 –15 0 k = 5 –3 – –3 – –3 –1 0 Q 1 (x) Q 2 (x) (x – 1) is not a factor of g(x) (x – 3) is not a factor of g(x) (x – 5) is a factor of g(x) (x + 1) is a factor of Q 1 (x)

10/29/2010 Section 3.3 v Rational Zero Test Example (continued) Factor completely: f(x) = 3x 4 – 12x 3 – 24x x + 45 We now have: f(x) = 3g(x) = 3(x – 5)(x + 1)(x 2 – 3) Synthetic division on Q 2 (x) = x 2 – 3 shows that none of the possible rational zeros (  1,  3) are zeros of Q 2 (x) To find the zeros of Q 2 (x) use difference of squares and solve for x using the zero product property Thus Note that the last two zeros are irrational Polynomial Functions Q 2 (x) = x 2 – 3 = (x – 3 )(x + 3 ) = 0   (x – 3 )(x + 3 )   f(x) = 3(x – 5)(x + 1)

10/29/2010 Section 3.3 v Equations Each new function f(x) we define leads to a new type of equation to solve when we set f(x) = 0 and find the zeros Examples Find all real solutions of: 1. x 4 – 1 = 0 2. x 3 = x 3. x 4 – 5x = 0 4. x 6 – 19x 3 = 216 Polynomial Functions

10/29/2010 Section 3.3 v Problem Solving Use the figure to find the width W of the rectangle from its length and area A. Also determine W when x = 5 inches. A = WL = W(x 2 + 1) A = 3x 3 + 3x – 5x 2 – 5 = 3x(x 2 + 1) – 5(x 2 + 1) = (3x – 5)(x 2 + 1) Thus At x = 5 inches W = 3(5) – 5 = 10 inches Note: To find W we could have used conventional long division Polynomial Functions W A = 3x 3 – 5x 2 + 3x – 5 x W = A = (3x – 5)(x 2 + 1) = 3x – 5

10/29/2010 Section 3.3 v Think about it !