Questions and Answers #6 Polynomials

Questions and Answers Sheet 6 Polynomials Question #1 Determine if the following set of polynomials fors a basis of 𝑃3 𝑝1 = 3 + 7𝑡, 𝑝2 = 5 + 𝑡 − 2𝑡 3 , 𝑝3 = 𝑡 − 2𝑡 2 , 𝑝4 = 1 + 16𝑡 − 6𝑡 2 + 2𝑡 3 Answer: We have given the polynomials 𝑝1 = 3 + 7𝑡, 𝑝2 = 5 + 𝑡 − 2𝑡 3 , 𝑝3 = 𝑡 − 2𝑡 2 , 𝑝4 = 1 + 16𝑡 − 6𝑡 2 + 2𝑡 3 The coordinate mapping gives the following coordinate vectors 3  5   0  7   1   1    ,   ,   and  0   0   −2         0   −2   0  1 16    . Write the above vectors as a column matrix and reduce the matrix into  −6    2 reduced row echelon form as follows. 3 5 0 1  3 5 0 1  7 1 1 16  1 −9 1 14   →   0 0 −2 −6  0 0 1 3       0 −2 0 2  0 1 0 −1  3 5 0 1  1 −9 1 14  1 −9 1 14   3 5 0 1   →  0 0 1 3  0 0 1 3      0 1 0 −1 0 1 0 −1  1 −9 3 5  0 0  0 1 1 3  0  0 0 5 0 1 1 0  0  0 0 5 0 1 1 0  0  0 0 1 0 1 1 14  1 0 0 1   3 5 → 1 3  0 0   0 −1 0 1 0 2  1 0 0 1   0 5 → 1 3  0 0   0 −1 0 1 0 2  1 0 −5 0 → 1 3  0   0 −1 0 0 2  1 0 −1 0 → 1 3  0   0 −1 0 0 1 0 1 0 1 0 0 0 2 0 1  1 3  0 −1 0 2 0 −5 1 3  0 −1 0 2 0 −1 1 3  0 −1 0 2 0 −1 1 3  0 0 From the above matrix, the column vectors are linearly dependent and do not form a basis for 𝑅4 Since the vector space 𝑅4 is isomorphic to 𝑃3 , hence the given set of polynomials does not form a basis of 𝑃3 (𝑅). Question #2 Find the indicated products. Assume all variables that appear as exponents represent positive integers. (𝑥 2𝑎 − 7)(𝑥 2𝑎 − 3) Answer: The given expression is, (𝑥 2𝑎 − 7)(𝑥 2𝑎 − 3) To find the product of the two polynomials, we need to distribute each term of the first polynomial to every term of the second polynomial. Then grouping together the like terms, we get the final product of the multiplication. On multiplying the given two polynomials, we get (𝑥 2𝑎 − 7)(𝑥 2𝑎 − 3) = 𝑥 2𝑎 (𝑥 2𝑎 − 3) − 7(𝑥 2𝑎 − 3) = 𝑥 2𝑎 ⋅ 𝑥 2𝑎 − 𝑥 2𝑎 ⋅ 3 − 7 ⋅ 𝑥 2𝑎 + 21 = 𝑥 2𝑎+2𝑎 + (−3 − 7)𝑥 2𝑎 + 21 = 𝑥 4𝑎 − 10𝑥 2𝑎 + 21 Therefore, the product of the given expression is 𝑥 4𝑎 − 10𝑥 2𝑎 + 21 Question #3 1 a) Find quadratic and cubic Taylor polynomials for 𝑥 3 at x=1 1 b) Show that these polynomials are upper and lower bounds for 𝑥 3 for x>1 1 c) Use these bounds to approximate 1.23 and 2 1 3 Answer: Consider the given information. 1 a) Here, we have to calculate the cubic Taylor polynomial for 𝑓(𝑥) = 𝑥 3 at x=1 𝑓(𝑥) = ∑∞ 𝑛=0 𝑓 (𝑛) (𝑎) 𝑛! (𝑥 − 𝑎)𝑛 The quadratic polynomial is defined as, 𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′(𝑎) 𝑛! (𝑥 − 𝑎) + 𝑓 ′′ 2! (𝑥 − 𝑎)2 Now, find the derivative of the function at x=1 f(1)=1 1 −2 𝑓 ′ (𝑥) = 𝑥 3 𝑓 ′ (1) = 3 1 3 And, 2 −5 𝑓 ′′ (𝑥) = − 𝑥 3 𝑓 ′′ (1) = 2 9 9 Substitute the values and find the quadratic polynomial. 1 1 𝑓(𝑥) = 1 + (𝑥 − 1) − (𝑥 − 1)2 3 1 1 9 = − (𝑥 − 1)2 + (𝑥 − 1) + 1 9 3 Now, find the third derivative. 10 𝑓 3 (𝑥) = 8 𝑓 3 (1) = 27𝑥 3 10 27 The cubic polynomial is defined as, 𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′(𝑎) 𝑛! (𝑥 − 𝑎) + 𝑓 ′′(𝑎) 2! (𝑥 − 𝑎)2 + Put the values in the polynomial. 1 1 𝑓(𝑥) = 1 + (𝑥 − 1) − (𝑥 − 1)2 + = 5 81 3 3 9 1 1 5 81 𝑓 3 (𝑎) 3! (𝑥 − 𝑎)3 (𝑥 − 1)3 (𝑥 − 1) − (𝑥 − 1)2 + (𝑥 − 1) + 1 9 3 1 b) For the given condition x>1, the second degree polynomial 𝑃2 (𝑥) ≤ 𝑥 3 1 As, the leading term is − (𝑥 − 1)2 9 1 And, For the given condition x>1, the third degree polynomial 𝑃3 (𝑥) ≥ 𝑥 3 5 As, the leading term is (𝑥 − 1)3 81 Thus, there are upper and lower bounds in the polynomials. c) Now, put the given values in the polynomials For x=1,2 1 1 𝑃2 (1.2) = − (1.2 − 1)2 + (1.2 − 1) + 1 9 3 ≈ 1.06222 And, 5 1 1 (1.2 − 1) − (𝑥 − 1)2 + (𝑥 − 1) + 1 𝑃3 (1.2) = 81 9 3 ≈ 1.07456 1 Thus, 1.0622 ≥ (1.2)3 ≥ 1.07456 For x=2 1 1 𝑃2 (2) = − (2 − 1)2 + (2 − 1) + 1 9 3 ≈ 1.223 And, 5 1 1 𝑃3 (2) = (2 − 1) − (2 − 1)2 + (2 − 1) + 1 81 9 3 ≈ 1.284 Thus, 1 1.223 ≤ 23 ≤ 1.284 Question #4 If 𝑓(𝑥) = 2𝑥+𝐵 𝑥−3 and f(5)=8, what is the value of B Answer: A function is a relation that maps each value in its domain to exactly one value in its range. A rational function is a function that is a ratio of two polynomials. The given function is a rational function as it is a ratio of two degree-one polynomials. The value of a function at a point are obtained by substituting the point in the function definition. 2𝑥+𝐵 The given function definition is 𝑓(𝑥) = and f(5)=8. Hence, substitute x=5 in function definition and 𝑥−3 equate it to 8 2𝑥+𝐵 𝑓(𝑥) = 𝑥−3 f(5)=8 2⋅5+𝐵 8= 8= 5−3 10+𝐵 2 10+B=16 10+B-10=16-10 B=6 Hence, the value of B is equal to 6. Question #5 Factor completely, or state that the polynomial is '. 𝑥 2 − 2𝑥 + 4 Answer: Consider the polynomials: 𝑥 2 − 2𝑥 + 4 To factorize the given polynomials by using completing square method, Add and substract by (1)2 in given polynomials, 𝑥 2 − 2𝑥 + +4 = 𝑥 2 − 2𝑥 + 4 + (1)2 − (1)2 = (𝑥 − 1)2 + 3 Further solving it, (𝑥 − 1)2 + 3 = 0 (𝑥 − 1)2 = −3 𝑥 − 1 = ±√−3 𝑥 − 1 = ±𝑖√3 𝑥 = 1 ± 𝑖√3 𝑥 = 1 + 𝑖√3 and 𝑥 = 1 − 𝑖√3 Question #6 Factor the following polynomials. a) 𝑥 2 − 8𝑥 b) 𝑥 3 + 3𝑥 2 − 18𝑥 c) 𝑥 3 − 10𝑥 Answer: Consider the polynomials a) 𝑥 2 − 8𝑥 b) 𝑥 3 + 3𝑥 2 − 18𝑥 c) 𝑥 3 − 10𝑥 a) 𝑥 2 − 8𝑥 𝑥 2 − 8𝑥 = 𝑥(𝑥 − 8) b) 𝑥 3 + 3𝑥 2 − 18𝑥 𝑥 3 + 3𝑥 2 − 18𝑥 = 𝑥(𝑥 2 + 3𝑥 − 18) = 𝑥(𝑥 2 + 6𝑥 − 3𝑥 − 18) = 𝑥(𝑥(𝑥 + 6) − 3(𝑥 + 6)) = 𝑥(𝑥 + 6)(𝑥 − 3) 𝑥 3 + 3𝑥 2 − 18𝑥 = 𝑥(𝑥 + 6)(𝑥 − 3) c) 𝑥 3 − 10𝑥 𝑥 3 − 10𝑥 = 𝑥(𝑥 2 − 10) Question #7 Explain the step by step of this factoring polynomials. 8𝑥 3 + 1 This problem is the sum of 2 perfect cubes. 8𝑥 3 + 1 = (2𝑥)3 + (1)3 and so we know that it is the fourth special form from above. Here is the factor for this polynomials. 8𝑥 3 + 1 = (2𝑥 + 1)(4𝑥 2 − 2𝑥 + 1) Answer: Given polynomial is, 8𝑥 3 + 1 The objective is to factorize the polynomial. The polynomial 8𝑥 3 + 1 is written as, (2𝑥)3 + 13 Now, Apply the sum of cubes formula, 𝑥 3 + 𝑦 3 = (𝑥 + 𝑦)(𝑥 2 − 𝑥𝑦 + 𝑦 2 ), by taking 2x as x and 1 as y. 8𝑥 3 + 1 = (2𝑥)3 + 13 = (2𝑥 + 1)(22 𝑥 2 − 2𝑥 + 1) This is the required answer. Question #8 Substract 9𝑥 2 𝑦 2 − 3𝑥 2 − 5 from the sum of −6𝑥 2 𝑦 2 − 𝑥 2 − 1 and 5𝑥 2 𝑦 2 + 2𝑥 2 − 1 Answer: First, find the sum of the polynomials −6𝑥 2 𝑦 2 − 𝑥 2 − 1 and 5𝑥 2 𝑦 2 + 2𝑥 2 − 1 Then, (−6𝑥 2 𝑦 2 − 𝑥 2 − 1) + (5𝑥 2 𝑦 2 + 2𝑥 2 − 1) = −6𝑥 2 𝑦 2 + 5𝑥 2 𝑦 2 − 𝑥 2 + 2𝑥 2 − 1 − 1 = (−6 + 5)𝑥 2 𝑦 2 + (−1 + 2)𝑥 2 − 2 = −𝑥 2 𝑦 2 + 𝑥 2 − 2 Therefore, the sum of the polynomials −6𝑥 2 𝑦 2 − 𝑥 2 − 1 and 5𝑥 2 𝑦 2 + 2𝑥 2 − 1 is −𝑥 2 𝑦 2 + 𝑥 2 − 2 Now, subtract 9𝑥 2 𝑦 2 − 3𝑥 2 − 5 from −𝑥 2 𝑦 2 + 𝑥 2 − 2 So, (−𝑥 2 𝑦 2 + 𝑥 2 − 2) − (9𝑥 2 𝑦 2 − 3𝑥 2 − 5) = −𝑥 2 𝑦 2 + 𝑥 2 − 2 − 9𝑥 2 𝑦 2 + 3𝑥 2 + 5 = −𝑥 2 𝑦 2 − 9𝑥 2 𝑦 2 + 𝑥 2 + 3𝑥 2 − 2 + 5 = (−1 − 9)𝑥 2 𝑦 2 + (1 + 3)𝑥 2 + 5 − 2 = −10𝑥 2 𝑦 2 + 4𝑥 2 + 3 Hence, the required polynomial is −10𝑥 2 𝑦 2 + 4𝑥 2 + 3 Question #9 For the function defined as follows, find the Taylor polynomials of degree 4 at 0. 1 𝑓(𝑥) = 𝑥−1 Answer: Taylor polynomials series: 𝑝4 (𝑥) = 𝑓(0) + 𝑓 1 (0) 1! 𝑥+ 𝑓 2 (0) 2! 𝑥2 + 𝑓 3 (0) 3! 𝑥3 + 𝑓 4 (0) 4! This is the taylors polynomial of degree 4 at 0. So we have function 1 𝑓(𝑥) = 𝑓(0) = 𝑓 1 (𝑥) 𝑥−1 1 = −1 0−1 −1 = (𝑥−1)2 −1 𝑓 1 (0) = (0−1)2 = −1 2 𝑓 2 (𝑥) = (𝑥−1)3 2 𝑓 2 (0) = (0−1)3 = −2 −6 𝑓 3 (𝑥) = (𝑥−1)4 −6 𝑓 3 (0) = (0−1)4 = −6 24 𝑓 4 (𝑥) = (𝑥−1)5 24 𝑓 4 (0) = (0−1)5 = −24 Putting the values in taylor polynomials 𝑥4 … after putting the values in equation 1 𝑝4 (𝑥) = −1 − 𝑥 − 𝑥 2 − 𝑥 3 − 𝑥 4 We get the this series of Taylors polynomials. Question #10 Determine whether the first polynomial can be expressed as a linear combination of other two polynomials. 2𝑥 3 − 2𝑥 2 + 12𝑥 − 6, 𝑥 3 − 2𝑥 2 − 5𝑥 − 3,3𝑥 3 − 5𝑥 2 − 4𝑥 − 9 Answer: The given polynomials are 2𝑥 3 − 2𝑥 2 + 12𝑥 − 6 𝑥 3 − 2𝑥 2 − 5𝑥 − 3 3𝑥 3 − 5𝑥 2 − 4𝑥 − 9 The polynomial can be written as vector form 2 1 3  −2   −2   −5 v1 =   , v2 =   , v2 =   12   −5  −4         −6   −3  −9  Let, 𝑣1 = 𝑎𝑣2 + 𝑏𝑣3 2 1 3  −2   −2      = a   + b  −5 12   −5  −4         −6   −3  −9   2   a + 3b   −2   −2a − 5b   =  12   5a − 4b       −6   −3a − 9b  ⇒ 𝑎 + 3𝑏 = 2 ⇒ −5𝑎 − 4𝑏 = 12 ⇒ −2𝑎 − 5𝑏 = −2 ⇒ −3𝑎 − 9𝑏 = −6 using equation 6b-5b=4-2 b=2 a=-4 These values of a, b are also santisty, hence polynomial having linear relationship 𝑃1 = −4𝑃2 + 2𝑃3 or 2𝑃3 = 𝑃1 + 4𝑃2 𝑃1 = I polynomial 𝑃2 = II polynomial 𝑃3 = III polynomial