University:
Rice University
Course:
MATH 102 | Single Variable Calculus II
Academic year:

2022
Views:

21

Pages:

3
Author:

MathMonk

Questions and Answers Sheet 19 Polynomials Question #1 Multiply, please: (𝑥 + ℎ)3 Answer: Apply perfect cube formula: (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2 𝑏 + 3𝑎𝑏 2 + 𝑏 (3) And we have: (𝑥 + ℎ)3 = 𝑥 3 + 3𝑥 2 ℎ + 3𝑥ℎ2 + ℎ(3) Question #2 Factor, please, 𝑦 = 4𝑥 2 + 17𝑥 + 4 Answer: Find a pair of factors of 𝐴𝐵 = 4 × 4 with sum C=17. We have the pair 16 and 1, so use this to split the middle term and factor by grouping: 𝑦 = 4𝑥 2 + 17𝑥 + 4 = (4𝑥 2 + 16𝑥) + (𝑥 + 4) = 4𝑥(𝑥 + 4) + 1(𝑥 + 4) = (4𝑥 + 1)(𝑥 + 4) Question #3 Factor 2𝑥 2 + 𝑥 − 1 Answer: 2𝑥 2 + 𝑥 − 1 Multiply the coefficient of the first term by the constant 2 × (−1) = −2 Rewrite, splitting the middle term x into 2x and -x 2𝑥 2 + 2𝑥 − 𝑥 = 1 Factor out the common terms 2𝑥(𝑥 + 1) − (𝑥 + 1)𝑟𝑖𝑔ℎ𝑡 ⇒ −(𝑥 + 1) = −1(𝑥 + 1) (𝑥 + 1)(2𝑥 − 1) Question #4 What is 15𝑥 2 + 17𝑥 + 4 = 0 in factor form? Answer: 15𝑥 2 + 17𝑥 + 4 = 0 15𝑥 2 + 5𝑥 + 12𝑥 + 4 = 0 5𝑥 × 3𝑥 + 5𝑥 × 1 + 12𝑥 + 4 = 0 5𝑥(3𝑥 + 1) + 12𝑥 + 4 = 0 5𝑥(3𝑥 + 1) + 4 × 3𝑥 + 4 × 1 = 0 5𝑥(3𝑥 + 1) + 4(3𝑥 + 1) = 0 (5𝑥 + 4)(3𝑥 + 1) = 0 Now, to solve for x, make each of these factors equal to 0 to find out what the x value will be: 5x+4=0 5x=-4 4 𝑥=− 5 And 3x+1=0 3x=-1 1 𝑥=− 3 Question #5 Solve 5𝑥 3 − 3𝑥 2 − 20𝑥 + 12 = 0 Answer: We have 5𝑥 3 − 3𝑥 2 − 20𝑥 + 12 = 0𝑍 𝑥 2 (5𝑥 − 3) − 4(5𝑥 − 3) = 0 (𝑥 2 − 4)(5𝑥 − 3) = 0 (𝑥 + 2)(𝑥 − 2)(5𝑥 − 3) = 0 3 Thus, 𝑥 = ±2, 5 Question #6 Write a polynomial in standard form, then classify it by degree and number of terms 3𝑥 2 − 6𝑥 4 + 2 + 4𝑥 Answer: This is the expression is a polynomial of fourth degree with 4 terms 1. 𝑥 4 with a coefficient -6 2. 𝑥 2 with a coefficient 3 3. x with a coefficient 4 4. And a free term 2 (without any variable) And the standart form is −6𝑥 4 + 3𝑥 2 + 4𝑥 + 2 Question #7 4-th derivative of (1+𝑥+𝑥 2 ) (1−𝑥+𝑥 2 ) using Taylor polynomial for 1 1−𝑥 Answer: 𝑓2 (𝑥) = = = = 1+𝑥+𝑥 2 1−𝑥+𝑥 2 (1+𝑥+𝑥 2 )(1+𝑥) (1−𝑥+𝑥 2 )(1+𝑥) (1+𝑥+𝑥 2 )(1+𝑥) 1+𝑥 3 1+2𝑥+2𝑥 2 +𝑥 3 1+𝑥 3 1+𝑥 =1+2 =1+2 1+𝑥 3 1 1+𝑥 3 + 2𝑥 1 1+𝑥 3 Question #8 Why must 𝑥 2 + 𝑥 + 1 be a factor of 𝑥 5 + 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1? Answer: It depends on what you count as telling. It follows from an easy, basic manipulation of familiar identities, but I don’t see any way to recognize this without appealing to something at least somewhat computational. I look at the sum 𝑥 5 + 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1 and immediately think of 𝑥 6 − 1, and from there it all falls out: 𝑥5 + 𝑥4 + 𝑥3 + 𝑥2 + 𝑥 + 1 = 2 = = 𝑥 6 −1 𝑥−1 (𝑥 3 ) −1 𝑥−1 (𝑥 3 −1)(𝑥 3 +1) 𝑥−1 = (𝑥 3 + 1) 𝑥 3 −1 𝑥−1 = (𝑥 3 + 1)(𝑥 2 + 𝑥 + 1) all using standard factorizations. Question #9 Why is 𝑥 3 − 5𝑥 injective on the rationals? Answer: If 𝑥 3 − 5𝑥 = 𝑦 3 − 5𝑦 with 𝑥 ≠ 𝑦 then 𝑥 3 − 𝑦 3 = 5(𝑥 − 𝑦) or 𝑥 2 + 𝑥𝑦 + 𝑦 2 = 5 Solving for x in terms of y: 𝑥= −𝑦±√𝑦 2 −4(𝑦 2 −5) 2 So 20 − 3𝑦 2 has to be a square of a rational number. This is equivalent to finding integer solutions to 20𝑞 2 − 3𝑝2 = 𝑛2 with (𝑝, 𝑞) = 1. Show that this isn't possible (mod 5) Question #10 Compute the product of the nonreal roots of the equation 𝑥 4 + 4𝑥 3 + 6𝑥 2 + 1004𝑥 + 1001 = 0 Answer: We are solving the equation 𝑥 4 + 4𝑥 3 + 6𝑥 2 + 1004𝑥 + 1001 = 0 Note that (𝑥 + 1)4 = 𝑥 4 + 4𝑥 3 + 6𝑥 2 + 4𝑥 + 1 So we are looking at (𝑥 + 1)4 + 1000𝑥 + 1000 = 0 Let 𝑦 = 𝑥 + 1. Our equation can be rewritten as 𝑦 4 + 1000𝑦 = 0. The real roots are y=0 and y=-10, and the other two roots are non-real. Now it's essentially over.