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Polynomial Factors Polynomial Factor Theorem

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7/23/2013 Polynomial Factors 2 The Remainder Theorem Polynomial f(x) divided by x – k yields a remainder of f(k) Question: Polynomial Factor Theorem = (x – k)Q(x) … leading to … 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)

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7/23/2013 Polynomial Factors 3 The Factor Theorem Polynomial f(x) has factor x – k if and only if f(k) = 0 Factor Theorem Corollary If (x – k) is a factor of f(x) = a n x n + a n–1 x n–1 + + a 1 x + a 0 then k is a factor of a 0 Polynomial Factor Theorem Question: What about the converse ?

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7/23/2013 Polynomial Factors 4 Example: f(x) = x 4 + x 3 – 19x 2 + 11x + 30 642 138 894 Polynomial Factor Theorem 1 1 –19 11 30 2 1 2 3 6 –13 –26 –15 –30 0 (x – 2) is a factor of f(x) Try factor k = 6... and 2 is a factor of 30 1 1 –19 11 30 1 723 149 924 (x – 6) NOT a factor of f(x)... BUT 6 IS a factor of 30 Try factor k = 2 Question: What if k = ±1, –2, ±3, ±5, –6, ±10, ±15 ? What about other factors of 30 ? ≠ 0 6

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7/23/2013 Polynomial Factors 5 Factor Theorem Example Given the graph of polynomial f(x) Estimate the degree of f(x) Even or odd degree ? Polynomial Factor Theorem x y ● ● ● (–7, 0) (2, 0) (11, 0) Odd degree ! WHY ?

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7/23/2013 Polynomial Factors 6 Factor Theorem Example Given the graph of polynomial f(x) 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 + 154 Polynomial Factor Theorem x y ● ● ● (–7, 0) (2, 0) (11, 0)

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7/23/2013 Polynomial Factors 7 Factor Theorem Example The graph of polynomial f(x) Question: Is f(x) = (x + 7)(x – 2)(x – 11) ? Polynomial Factor Theorem x y ● ● ● (–7, 0) (2, 0) (11, 0) Not necessarily ! f(x) = (x + 7)Q 1 (x) = (x + 7)(x – 2)Q 2 (x) = (x + 7)(x – 2)(x – 1)Q 3 (x)

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7/23/2013 Polynomial Factors 8 Factor Theorem Example The graph of polynomial f(x) Question: Polynomial Factor Theorem x y ● ● ● (–7, 0) (2, 0) (11, 0) f(x) = (x + 7)(x – 2)(x – 1)Q 3 (x) What is Q 3 (x) ? Is Q 3 (x) constant ? Does Q 3 (x) have factors x – k ?

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7/23/2013 Polynomial Factors 9 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) Polynomial Factor Theorem x y (–3, 0)(5, 0) ● ● Note: (x + 3)(x – 5) = x 2 – 2x 2 – 15 f(–3) = f(5) = 0 so(x + 3) and (x – 5) are factors WHY !

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7/23/2013 Polynomial Factors 10 Factor Theorem Example Given the graph of f(x) Polynomial Factor Theorem x y (–3, 0)(5, 0) ● ● Is f(x) equal to x 2 – 2x – 15 ? Probably not ! Question: (x + 3)(x – 5) = x 2 – 2x 2 – 15 What is the graph of x 2 – 2x – 15 ? Note:f(x) = (x + 3)Q 1 (x) = (x + 3)(x – 5)Q 2 (x) Which ones and how many ? Does Q 2 (x) have factors (x – k) ?

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7/23/2013 Polynomial Factors 11 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) Intercepts, Zeros and Factors WHY ?

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7/23/2013 Polynomial Factors 12 Example: Completely Factor f(x) = 2x 4 + 14x 3 + 18x 2 – 54x – 108 Now use synthetic division to check out zeros Completely Factored Polynomials = 2(x 4 + 7x 3 + 9x 2 – 27x – 54)

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7/23/2013 Polynomial Factors 13 Completely Factored Polynomials f(x) = 2(x 4 + 7x 3 + 9x 2 – 27x – 54) 1 –3 4 –12 9 –18 54 0 1 7 9 –27 –54 –3 x – k = x – (–3) (x + 3) is a factor 1 –3 1 18 0 1 4 –3 –18 –6 (x + 3) is a second factor Q1Q1 –3

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7/23/2013 Polynomial Factors 14 Completely Factored Polynomials f(x) = 2(x 4 + 7x 3 + 9x 2 – 27x – 54) 1 7 9 –27 –54 0 1 4 –18 –3 (x + 3) is a factor 1 1 0–6 (x + 3) is a second factor 1 –3 –2 6 0 1 1 –6 (x + 3) is a third factor (x – 2) is a fourth factor 2(x + 3) 3 (x – 2) Note: x = –3 is a repeated zero of multiplicity 3 Q2Q2 –3

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7/23/2013 Polynomial Factors 15 Completely Factored Polynomials f(x) = 2(x 4 + 7x 3 + 9x 2 – 27x – 54) 1 –3 4 –12 9 –18 54 0 1 7 9 –27 –54 –3 x – k = x – (–3) (x + 3) is a factor 1 –3 1 18 0 1 4 –3 –18 –6 (x + 3) is a second factor 1 –3 –2 6 0 1 1 –6 (x + 3) is a third factor (x – 2) is a fourth factor 2(x + 3) 3 (x – 2) Note: x = –3 is a repeated zero of multiplicity 3 Q1Q1 Q2Q2 –3

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7/23/2013 Polynomial Factors 16 Complete Factoring with Multiple Zeros General Form f(x) = a n x n + a n–1 x n–1 +... + a 1 x + a 0 = a n (x – k n )(x – k n–1 )... (x – k 1 ) Some of the zeros k i may be repeated Polynomial Functions

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7/23/2013 Polynomial Factors 17 Complete Factoring with Multiple Zeros f(x) = a n (x – k n )(x – k n–1 )... (x – k 1 ) Degree of f(x) = n Number of zeros is m, m ≤ n Real zeros occur at x-intercepts Counting multiplicities, total number of zeros is n Polynomial Functions

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7/23/2013 Polynomial Factors 18 Complete Factoring with Multiple Zeros Example f(x) = 2(x + 3) 3 (x – 2) deg f(x) = 4 One zero at 2, one at –3 of multiplicity 3 Total zeros, counting multiplicities, 1 + 3 = 4 = deg f(x) Polynomial Functions

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7/23/2013 Polynomial Factors 19 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) Polynomial Functions Zeros of f(x) are –1, 0, 3

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7/23/2013 Polynomial Factors 20 Complete Factoring Examples 2. Given: f(x) is a quadratic polynomial lead coefficient is 7 f(–3) = 0 and f(2) = 0 Write f(x) in completely factored form Polynomial Functions

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7/23/2013 Polynomial Factors 21 Complete Factoring Examples Write f(x) in completely 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

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7/23/2013 Polynomial Factors 22 Even/Odd Multiplicity 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 ● ●

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

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