Polynomial factorization #5 - Questions and Answers

Question 1 Find a degree 3 polynomial having zeros -8, 2, and 6 and the coefficient of x³ equal 1. the polynomial is: Answer Step 1 Given: Zeroes of the degree 3 polynomial: -8, 2 and 6 Step 2 Then the factors will be (x+8),(x-2) and (x-6) Step 3 The degree 3 equation is given below: (x+8)(x-2)(x-6) (x² - 2x + 8x-16) (x – 6) (x²+ 6x-16) (x – 6) (x-6) (x² + 6x - 16) x³+6x²-16x-6x²-36x + 96 x³ - 52x + 96 Step 4 The 3 degree polynomial with zeros -8, 2 and 6 and coefficient x³ equal to 1 is: x³- 52x + 96 Question 2 Find the Taylor polynomial T₃(x) for the function f centered at the number a. f(x) = x + e⁻ˣ, a = 0 T₃(x) =? Answer f(x) = x + e⁻ˣ f(0) = 1 f'(x) = 1 - e⁻ˣ f'(0) = 0 f"(x) = e⁻ˣ f"(0) = 1 f'''(x) = -e⁻ˣ f'''(0) = -1 The Taylor polynomial to the 3rd degree is 1 + x²/2 - x³/6 Graph x + e⁻ˣ and 1 + x²/2 - x³/6 on your calculator and the y-intercepts should be same Question 3 Find all real solutions of the equation by factoring. 6x(x-1)=21-x Answer 1 Step 1 We have to find all real solutions of the equation by factorizing: 6x(x-1)=21-x Solving given equation, 6x(x-1)=21-x 6x² - 6x = 21 - x 6x² - 6x + x = 21 6x² - 5x - 21 = 0 Now solving by factorization, 6x² - 14x + 9x - 21 = 0 2x(3x-7)+3(3x-7)=0 (3x-7)(2x+3)=0 Step 2 Either, 3x-7=0 3x=7 x = 7/3 or, 2x+3=0 2x=-3 x = -3/2 Hence, solutions of the equation are x = -3/2, 7/3 Answer 2 The given quadratic equation is 6x(x-1)=21-x 6x(x-1)=21-x [given equation] 6x² - 6x = 21 - x 6x² - 6x + x = 21 - x + x 6x² - 5x = 21 6x² - 5x - 21 = 21 - 21 6x² - 5x - 21 = 0 [quadratic form] 6x² + 9x - 14x - 21 = 0 3x(2x+3)-7(2x+3)=0 (3x-7)(2x+3)=0 [factor] 3x-7=0 or 3x+3=0 [zero-product property] x = 7/3 or x = -3/2 The solutions are x = 7/3 and x = -3/2 Question 4 Complete Factorization Factor the polynomial completely, and find all its zeros.State the multiplicity of each zero. P(x) = x⁵ + 7x³ Answer Concept used: The multiplicity of zero of the polynomial having factor (x-c) that appears k times in the factorization of the polynomial is k. Calculation: The given polynomial is P(x) = x⁵ + 7x³. Factor the above polynomial to obtain the zeros. P(x) = x⁵ + 7x³ = x³(x²+7) = x³(x² - (√7i)²) = x³(x - √7i)(x + √7i) Substitute 0 for P (x) in the polynomial P(x) = x⁵ + 7x³ to obtain the zeros of the polynomial x³(x - √7i)(x + √7i) Further solve for the value of x as, x³ = 0, (x - √7i) = 0 and (x + √7i) = 0 x = 0, x = √7i and x = −√7i The zeros of the polynomial P(x) = x⁵ + 7x³ appears three times and one time in the polynomial therefore, the multiplicity of the zero 0 is 3, √7i and √-7i is 1. Conclusion: Thus, the factorization of the polynomial P(x) = x⁵ + 7x³ is P(x) = x³(x - √7i)(x + √7i), zeros of the polynomial are 0 and + √7i and the multiplicity of the zero 0 is 3, √7i and -√7i is 1. Question 5 Determine how many linear factors and zeros each polynomial function has. P(x) = x⁴⁰ + x³⁹ Answer Amount of zeros theorem: The degree three polynomial function P(x) holds true if numerous zeros are counted independently n(n > 0) has exactly n zeros among the complex numbers. The Polynomial Factorization Theorem: If n > 0 if P(x) has exactly n linear elements and P(x) is a polynomial function of the nth degree: P(x) = aₙ(x-c₁) (x-c₂) (x-c₃) ... (x-cₙ) Where c₁, c₂, c₃, ..., cₙ are numbers and aₙ, is the leading coefficient of P(x). Finding zeros: Given: P(x) = x⁴⁰ + x³⁹ The polynomial function P(x) has n zeros, according to the Number of Zeros Theorem. where n is the polynomial function's degree Here n = 40 Hence the zeros of the polynomial function P(x) = x⁴⁰ + x³⁹ is 40 Finding linear factors: Given: P(x) = x⁴⁰ + x³⁹ The polynomial function P(x) has n linear factors, according to The Polynomial Factorization Theorem. where n is the polynomial function's degree Here n = 40 Hence the linear factors of the polynomial function P(x) = x⁴⁰ + x³⁹ is 40 The polynomial function's number of linear elements is the conclusion. P(x) = x⁴⁰ + x³⁹ are 40. The zeros of the polynomial function P(x) = x⁴⁰ + x³⁹ are 40.