University:
North Central College
Course:
MATH 152 | Calculus II
Academic year:

2024
Views:

404

Pages:

2
Author:

nesikhiyawl0

Question 1 2x - 3x² + 4 - 5x⁶ Leading coefficient: Degree of a Polynomial: Terms of a Polynomial: Standard form: Answer Step 1 2x - 3x² + 4 - 5x⁶ Step 2 2x - 3x² + 4 - 5x⁶ Leading Coefficient: coefficient of the highest degree of the variable ie coefficient of x⁶: -5 Degree of a Polynomial: Highest degree of the polynomial : 6 Terms of a Polynomial: 4, because there are 4 terms in this polynomial Standard Form: Is arranging / writing the polynomial in the decreasing order of there degree: -5x⁶ - 3x⁴ + 2x + 4 Question 2 Use the linear factorization theorem to construct a polynomial function with the given condition. n=3 with real coefficients, zeros = -4, 2+3i. Answer Step 1 Given: f(x) is a polynomial with real coefficients of degree 3 Zeroes of f(x) are -4, 2 + 3i Since f(x) is a polynomial with real coefficients the complex roots exist as complex cojugates So, 2-3i is a zero of f(x) as 2+3i is a zero. Step 2 Hence, f(x)=(x+4)(x-(2+3i))(x-(2-3i)) f(x)=(x-(-4))((x-2)-3i)((x-2)+3i) f(x) = (x + 4) ((x - 2)² + 9) f(x) = (x + 4) (x² - 4x + 13) f(x) = x³ - 4x² + 13x + 4x² - 16x + 52 f(x) = x³ - 3x + 52 Hence, the polynomial is x³ - 3x + 52 Question 3 Why isn't factoring x⁴ - 81 as (x² + 9)(x² - 9) a complete factorization? Answer Step 1 Given statement is x⁴ - 81 as (x²+9) (x² - 9) is complete factorization. Step 2 (x² - 9) factors are also possible. ∴ (x² + 9) (x²-9) is not complete factorization of x⁴ – 81. Step 3 Factors of (x² - 9) are (x² - 9) = (x + 3)(x - 3) ∴ Complete factorization of x⁴ – 81 is (x²+9)(x + 3)(x - 3). Question 4 Find the prime factorization of 27. Answer Step 1 Factors of 27: 1, 3, 9, 7 Step 2 But only 3 is a prime number. Step 3 So prime factorization of 27, 27 = 3 x 3 x 3 27 = 3³ Question 5 Find the prime factorization of 76 Answer Step 1 We have to find prime factorization. Step 2 76 = 2 × 38 = 2 × 2 × 19 76 = 2² × 19¹ Which is prime factorization of 76.