# 4.1: Polynomial Functions

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4.1: Polynomial Functions
A polynomial is an algebraic function combining coefficients variables (with positive integers as exponents) and constants. Any letter may be used as the variable Some examples: x3 – 6x2 + ½ y15 + y10 + 7 w – 6.7 12 A polynomial that consists of only a constant term is called a constant polynomial. The zero polynomial is the constant polynomial 0.

4.1: Polynomial Functions
Degree of a polynomial: The exponent of the highest power Leading coefficient The coefficient of the highest power Examples Polynomial Degree Leading coefficient Constant term 6x7 + 4x3 + 5x2 – 7x + 10 7 6 10 x3 3 1 12 0x9 + 2x6 + 3x7 + x8 – 2x – 4 8 -4

4.1: Polynomial Functions
Polynomial Division – long division Divide 3x4 – 8x2 – 11x + 1 by x – 2.

4.1: Polynomial Functions
Synthetic Division Only works when dividing by a first degree polynomial (e.g. “x – 2”, “x + 7”) List the divisor separate from the dividend Bring down the opposite of the first term Multiply the first term by the divisor, place under the next coefficient Add terms and repeat You answer begins with the first coefficient, using one exponent less than the dividend. The last number you wind up with is the remainder

4.1: Polynomial Functions
Divide x5 + 5x4 + 6x3 – x2 + 4x + 29 by x + 3 The answer is x4 + 2x3 – 1x + 7, R:8

4.1: Polynomial Functions
Checking polynomial division Dividend = Divisor • Quotient + Remainder In English? Equation divided = dividing equation • answer + remainder Checking the previous problem Dividend: x5 + 5x4 + 6x3 – x2 + 4x + 29 Divisor: x + 3 Quotient: x4 + 2x3 – 1x + 7 Remainder: 8 (x4 + 2x3 – 1x + 7)(x + 3) + 8 x5 + 2x4 – 1x2 + 7x + 3x4 + 6x3 – 3x x5 + 5x4 + 6x3 – x2 + 4x + 29

4.1 Polynomial Functions If the remainder after division is 0, then the both the divisor (dividing function) and quotient (answer) are factors of the dividend Determine if 2x2 + 1 is a factor of 6x3 – 4x2 + 3x – 2

4.1 Polynomial Functions Assignment Page 248-249 Problems 1-25, odd
Show work (obviously) For problems 19-25, you will have to use long division You won’t get credit without doing (I already know that the work from 9-15 is in the back of the book)

Chapter 4: Polynomial and Rational Functions 4.1: Polynomial Functions
Essential Question: Learning? Wha…?

4.1 Polynomial Functions Remainder Theorem
When a polynomial is divided by a first-degree polynomial, such as x – 3 or x + 5, the remainder is a constant. For example, (x3 – 2x2 – 4x + 5) / (x – 3) = x2 + x – 1, R: 2 If you were to plug in the opposite of the divisor (“3”… like using synthetic division), you’ll get the remainder. (3)3 – 2(3)2 – 4(3) + 5 = 27 – 18 – = 2 If a polynomial f(x) is divided by x – c, the remainder is f(c)

4.1 Polynomial Functions Example #1 Example #2
Find the remainder when x79 + 3x is divided by x – 1 Example #2 Find the remainder when 3x4 – 8x2 + 11x + 1 is divided by x + 2 Note: Make sure you’re using parenthesis if you’re plugging into your calculator… (1)79 + 3(1) = = 9 3(-2)4 – 8(-2)2 + 11(-2) + 1 = 48 – 32 – = -5

4.1 Polynomial Functions Zeros and Factors Example
A polynomial function f(x) has a linear factor x – a if and only if f(a) = 0 In English: Rather than going through the time to do synthetic division, just plug in that opposite number. If the remainder is 0, then you’ve got a factor. Example Show that x – 3 is a factor of x3 – 4x2 + 2x + 3 by using the Factor Theorem. (3)3 – 4(3)2 + 2(3) + 3 = 27 – = 0

4.1 Polynomial Functions Fundamental Polynomial Connections
For f (x) = 15x3 – x2 – 114x + 72, find the following: The x-intercepts of the graph of f The zeros of f The solutions to 15x3 – x2 – 114x + 72 The linear factors with real coefficients of 15x3 – x2 – 114x + 72 Solution Graph the function. Use the root feature. Same as the roots… Same as the roots… Can be verified by substitution

4.1 Polynomial Functions Finishing that last part…
This isn’t the original polynomial [15x3 – x2 – 114x + 72] Note that if we multiplied the factors out, we’d have a leading term of x3, which means that our factors must be multiplied by 15

4.1 Polynomial Functions A Polynomial with Specific Zeros Solutions
Find three polynomials of different degrees that have 1, 2, 3, and -5 as zeros. Solutions A polynomial that has 1, 2, 3, and -5 as zeros must have (x – 1), (x – 2), (x – 3), and (x + 5) as factors. There are multiple solutions g(x) = (x – 1)(x – 2)(x – 3)(x + 5) h(x) = 8(x – 1)(x – 2)(x – 3)2(x + 5) k(x) = 2(x + 4)2(x – 1)(x – 2)(x – 3)(x + 5)(x2 + x + 1) Note that g has degree 4, h has degree 5 and k has degree 8

4.1 Polynomial Functions The Number of Zeros of a Polynomial Example
A polynomial of degree n has at most n distinct real zeros Example f(x) = 18x4 – 51x3 – 187x2 – 56x + 80 f(x) has at most 4 zeros

4.1 Polynomial Functions Assignment Page 249 Problems 27 – 53, odds