Chapter 5 Study Guide 5-1 Study Guide - Operations with Polynomials Multiply and Divide Monomials Negative exponents are a way of expressing the multiplicative inverse of a number. Negative Exponents = and = for any real number a ≠ 0 and any integer n. When you simplify an expression, you rewrite it without powers of powers, parentheses, or negative exponents. Each base appears only once, and all fractions are in simplest form. The following properties are useful when simplifying expressions. Product of Powers = for any real number a and integers m and n. Quotient of Powers = for any real number a ≠ 0 and integers m and n. Properties of Powers For a, b real numbers and m, n integers: = = = b ≠ 0 = or a ≠ 0,b ≠ 0 Example: Simplify. Assume that no variable equals 0. a. ( ( = 25 = 75 = = 75 b. = = = – Exercises: Simplify. Assume that no variable equals 0. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 4j(–)(3) 12. 5-1 Study Guide (continued) Operations with Polynomials Polynomial a monomial or a sum of monomials
Polynomial Function A polynomial function of degree n can be described by an equation of the form P(x) = + + … + + x + , where the coefficients , , , …, represent real numbers, is not zero, and n represents a nonnegative integer. Example 1: What are the degree and leading coefficient of 3 – 2 – 7 + ? Rewrite the expression so the powers of x are in decreasing –2 + + 3 – 7 This is a polynomial in one variable. The degree is 4, and the leading coefficient is –2. Example 2: Find f(–5) if f(x) = + 2 – 10x + 20. f(x) = + 2 – 10x + 20 Original function f(–5) = + – 10(–5) + x with –5. = –125 + 50 + 50 + 20 Evaluate. = –5 Simplify. Example 3: Find g( – 1) if g(x) = + 3x – 4. g(x) = + 3x – 4 Original function g( – 1) = + 3( – 1) – 4 Replace x with – 1. = – 2 + 1 + 3 – 3 – 4 Evaluate. = + – 6 Simplify. Exercises State the degree and leading coefficient of each polynomial in one variable. If it is not a polynomial in one variable, explain why. 1. 3 + 6 – + 12 2. 100 – 5 + 10 3. 4 + 6 + 8 – 10 + 20 4. 4 – 3xy + 16 5. 8 – 9 + 4 – 36 6. – + – Find f(2) and f(–5) for each function. 7. f(x) = – 9 8. f(x) = 4 – 3 + 2x – 1 9. f(x) = 9 – 4 + 5x + 7 5-3 Guide (continued) Polynomial Functions Graphs of Polynomial Functions End Behavior of Polynomial Functions If the degree is even and the leading coefficient is positive, then f(x) → +∞ as x → –∞ f(x) → +∞ as x → +∞ If the degree is even and the leading coefficient is negative, then f(x) → –∞ as x → –∞ f(x) → –∞ as x → +∞ If the degree is odd and the leading coefficient is positive, then
x f(x) –2 35 –1 –2 0 –5 1 –4 2 19 The changes in sign indicate that are zeros between x = –2 and x = –1 and between x = 1 and x = 2. Exercises Graph each function by making a table of values. Determine the values of x between which each real zero is located. 1. f(x) = – 2 + 1 2. f(x) = + 2 – 5 3. f(x) = – + 2 – 1 4. f(x) = – 3 + 4 5. f(x) = 3 + 2x – 1 6. f(x) = – 3 + 1 5-4 Study Guide and Intervention (continued) Analyzing Graphs of Polynomial Functions Maximum and Minimum Points A quadratic function has either a maximum or a minimum on its graph. For higher degree polynomial functions, you can find turning points, which represent relative maximum or relative minimum points. Example: Graph f(x) = + 6 – 3. Estimate the x-coordinates at which the relative maxima and minima occur. Make a of values and graph the function. x f(x) –5 22 –4 29 ← indicates a relative maximum –3 24 –2 13 –1 2 ← zero between x = –1, x = 0 0 –3 ←indicates a relative minimum 1 4 2 29 A relative maximum occurs at x = –4 and a relative minimum occurs at x = 0.
Exercises Solve each equation. 1. = 49 2. – 6 = –8 3. – 3 = 54 4. 3 – 48 = 0 5. – 16 + 64 = 0 6. – 5 + 4 = 0 7. – 29 + 100 = 0 8. 4 – 73 + 144 = 0 5-6 Study Guide and Intervention The Remainder and Factor Theorems Synthetic Substitution Remainder Theorem The remainder, when you divide the polynomial f (x ) by (x – a), is the constant f (a). f(x) = q(x) ⋅ (x – a) + f(a), where q(x) is a polynomial with degree one less than the degree of f(x). Example 1: If f(x) = 3 + 2 – 5 + x – 2, find f(–2). Method 1: Synthetic Substitution By the Remainder Theorem, f(–2) should be the remainder when you divide the polynomial by x + 2. –2 3 2 –5 1 –2 –6 8 –6 10 3 –4 3 –5 8 The remainder is 8, so f(–2) = 8. Method 2 Direct Substitution Replace x with –2. f(x) = 3 + 2– 5 + x – 2 f(–2) = 3 + 2– 5+ (–2) – 2 = 48 – 16 – 20 – 2 – 2 or 8 So f(–2) = 8 Example 2: If f(x) = 5+ 2x – 1, find f(3). Again, by the Remainder Theorem, f(3) should be the remainder when you divide the polynomial by x – 3. 3 5 0 2 –1
15 45 141 5 15 47 140 The remainder is 140, so f(3) = 140. Exercises Use synthetic substitution to find f(–5) and f for each function. 1. f(x) = –3 + 5x – 1 2. f(x) = 4 + 6x – 7 3. f(x) = – + 3 – 5 4. f(x) = + 11 – 1 Use synthetic substitution to find f(4) and f(–3) for each function. 5. f(x) = 2 + – 5x + 3 6. f(x) = 3 – 4x + 2 7. f(x) = 5– 4 + 2 8. f(x) = 2 – 4 + 3 + x – 6 9. f(x) = 5 + 3 – 4 – 2x + 4 10. f(x) = 3 – 2 – + 2x – 5 5-6 Guide (continued) Factors of Polynomials The Factor Theorem can help you find all the factors of a polynomial. Factor Theorem The binomial x – a is a factor of the polynomial f(x) if and only if f(a) = 0. Example: Show that x + 5 is a factor of + 2– 13x + 10. Then find the remaining factors of the polynomial. By the Factor Theorem, the binomial x + 5 is a factor of the polynomial if –5 is a zero of the polynomial function. To check this, use synthetic substitution. –5 1 2 –13 10 –5 15 –10 1 –3 2 0 Since the remainder is 0, x + 5 is a factor of the polynomial. The polynomial + 2 – 13x + 10 can be factored as (x + 5)( – 3x + 2). The depressed polynomial – 3x + 2 can be factored as (x – 2)(x – 1). So + 2 – 13x + 10 = (x + 5)(x – 2)(x – 1). Exercises Given a polynomial and one of its factors, find the remaining factors of the polynomial. 1. + – 10x + 8; x – 2 2. – 4 – 11x + 30; x + 3
So – 5 is another zero. Use the Quadratic Formula on the depressed polynomial – 2x + 4 to find the 1 zeros, 1 ± i . The function has two real zeros at 3 and –5 and two imaginary zeros at 1 ± i . Exercises Find all zeros of each function. 1. f(x) = + + 9x + 9 2. f(x) = – 3 + 4x – 12 3. p(a) = – 10 + 34a – 40 4. p(x) = – 5 + 11x – 15 5-8 Study Guide - Rational Zero Theorem Identify Rational Zeros Rational Zero Theorem Let f (x) = + + … + + x + represent a polynomial function with integral coefficients. If is a rational number in simplest form and is a zero of y = f(x), then p is a factor of and q is a factor of . Corollary (Integral Zero Theorem If the coefficients of a polynomial are integers such that = 1 and ≠ 0, any rational zeros of the function must be factors of . Example : List all of the possible rational zeros of each function. a. f(x) = 3 – 2 + 6x – 10 If is a rational root, then p is a factor of –10 and q is a factor of 3. The possible values for p are ±1, ±2, ±5, and ±10. The possible values for q are 61 and 63. So all of the possible rational zeros are = ±1, ±2, ±5, ±10, ± , ± , ± , and ± . b. q(x) = – 10 + 14x – 36 Since the coefficient of x3 is 1, the possible rational zeros must be the factors of the constant term –36. So the possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, and ±36. Exercises List all of the possible rational zeros of each function. 1. f(x) = + 3 – x + 8 2. g(x) = – 7 + 3 + x – 20 3. h(x) = – 7 – 4 + x – 49 4. p(x) = 2 – 5 + 8 + 3x – 5