Polynomial arithmetic #2 - Questions and Answers

Question 1 We need to find: The real zeros of the polynomial function form an arithmetic sequence: 𝑓 ( 𝑥 ) = 𝑥 4 − 4 𝑥 3 − 4 𝑥 2 + 16 𝑥 . f(x)=x 4 −4x 3 −4x 2 +16x. Answer: Concept: A rational expression is a fraction that is a quotient of two polynomials. A rational function is defined by two polynomial functions. The given polynomial function forms an arithmetic sequence: 𝑓 ( 𝑥 ) = 𝑥 4 − 4 𝑥 3 − 4 𝑥 2 + 16 𝑥 . f(x)=x 4 −4x 3 −4x 2 +16x. Here, the constant is 0. The above equation can be rewritten as: 𝑓 ( 𝑥 ) = 𝑥 ( 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16 ) . f(x)=x(x 3 −4x 2 −4x+16). The possibilities for 𝑥 x are 0 , ± 1 , ± 2 , ± 4 0,±1,±2,±4, and 8 8. Factoring 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16 x 3 −4x 2 −4x+16, we get: 𝑥 ( 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16 ) = ( 𝑥 + 2 ) 2 ( 𝑥 − 4 ) . x(x 3 −4x 2 −4x+16)=(x+2) 2 (x−4). Thus, the real zeros are − 2 , 0 , 2 , 4 −2,0,2,4. Question 2 The general term of a sequence is given as: 𝑎 𝑛 = 2 𝑛 . a n ​ =2 n . Determine whether the sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. Answer: If the subsequent terms in a series differ by a constant (common difference), the sequence is an arithmetic sequence. If there is a constant ratio between the terms in a series, the sequence is a geometric sequence. Substitute 𝑛 = 1 , 2 , 3 n=1,2,3: 𝑎 1 = 2 1 = 2 ,   𝑎 2 = 2 2 = 4 ,   𝑎 3 = 2 3 = 8. a 1 ​ =2 1 =2,a 2 ​ =2 2 =4,a 3 ​ =2 3 =8. The difference between terms is not constant, so the sequence is not arithmetic. The ratio of successive terms is constant ( 𝑎 2 𝑎 1 = 2 , 𝑎 3 𝑎 2 = 2 a 1 ​ a 2 ​ ​ =2, a 2 ​ a 3 ​ ​ =2), so the sequence is geometric with a common ratio of 2. Question 3 The general term of a sequence is given as: 𝑎 𝑛 = 𝑛 + 5. a n ​ =n+5. Determine whether the sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. Answer: Step 1: Given 𝑎 𝑛 = 𝑛 + 5 a n ​ =n+5. Step 2: Find the first three terms: 𝑎 1 = 1 + 5 = 6 ,   𝑎 2 = 2 + 5 = 7 ,   𝑎 3 = 3 + 5 = 8. a 1 ​ =1+5=6,a 2 ​ =2+5=7,a 3 ​ =3+5=8. The difference between successive terms is constant ( 7 − 6 = 1 ,   8 − 7 = 1 7−6=1,8−7=1), so the sequence is arithmetic with a common difference of 1. Question 4 Determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. The sequence is: 9 = 10 11 𝑛 . 9= 11 10 ​ n. Answer 1: Step 1: Since the expression is a linear function of 𝑛 n, it is an arithmetic sequence. The common difference is the coefficient of 𝑛 n, which is − 10 11 − 11 10 ​ . Step 2: To find the sum of the first 50 terms: 𝑆 𝑛 = 𝑛 2 [ 2 𝑎 + ( 𝑛 − 1 ) 𝑑 ] . S n ​ = 2 n ​ [2a+(n−1)d]. Plug in 𝑛 = 50 , 𝑎 = 9 n=50,a=9, and 𝑑 = − 10 11 d=− 11 10 ​ : 𝑆 50 = 50 2 [ 2 ( 9 ) + ( 50 − 1 ) ( − 10 11 ) ] . S 50 ​ = 2 50 ​ [2(9)+(50−1)(− 11 10 ​ )]. Simplifying: 𝑆 50 = 25 [ 18 − 45.45 ] = 25 ( − 27.45 ) = − 686.25. S 50 ​ =25[18−45.45]=25(−27.45)=−686.25. Question 5 Explain with numerical examples what Real Numbers and Algebraic Expressions are. Explain factoring and finding LCMs (least common multiples). Answer: Step 1: Real Numbers: Real numbers can be plotted on a real number line. Examples: Whole numbers: 4 , − 3 , 0 4,−3,0. Rational numbers: 3.5 , 2 7 3.5, 7 2 ​ . Irrational numbers: 2 , 𝜋 2 ​ ,π. Step 2: Algebraic Expressions: An algebraic expression consists of variables and constants. Examples: Simple: 2 𝑥 + 3 2x+3. Complex: ( 2 𝑥 2 + 1 ) 2 (2x 2 +1) 2 . Step 3: Factoring and LCMs: Factoring: Breaking an expression into simpler terms, e.g., 𝑥 2 − 9 = ( 𝑥 − 3 ) ( 𝑥 + 3 ) x 2 −9=(x−3)(x+3). LCMs: The smallest number divisible by given numbers. For 12 12 and 18 18, LCM is 36 36. Вы сказали: ChatGPT сказал: Extracted Text: Question 1 In each of the following, list three terms that continue the arithmetic or geometric sequences. Identify the sequences as arithmetic or geometric. Given sequences: 2 , 6 , 18 , 54 , 162 2,6,18,54,162 1 , 8 , 15 , 22 , 29 1,8,15,22,29 11 , 15 , 19 , 23 , 27 11,15,19,23,27 Answer: The given sequence is 2 , 6 , 18 , 54 , 162 2,6,18,54,162: The ratio of two consecutive terms is constant ( 𝑟 = 6 2 = 3 r= 2 6 ​ =3). Therefore, the sequence is geometric with a common ratio 𝑟 = 3 r=3. Next three terms: 486 , 1458 , 4374. 486,1458,4374. The given sequence is 1 , 8 , 15 , 22 , 29 1,8,15,22,29: The difference between two consecutive terms is constant ( 𝑑 = 8 − 1 = 7 d=8−1=7). Therefore, the sequence is arithmetic with a common difference 𝑑 = 7 d=7. Next three terms: 36 , 43 , 50. 36,43,50. The given sequence is 11 , 15 , 19 , 23 , 27 11,15,19,23,27: The difference between two consecutive terms is constant ( 𝑑 = 15 − 11 = 4 d=15−11=4). Therefore, the sequence is arithmetic with a common difference 𝑑 = 4 d=4. Next three terms: 31 , 35 , 39. 31,35,39. Question 2 (a) To calculate: The value of 𝑃 ( 2 ) P(2) without a calculator using both forms of the polynomial. The value of 𝑃 ( 2 ) P(2) is 236 236. (b) The least amount of arithmetic operations was performed in (c) using Horner's Method. Answer: (a) Given: A fourth-degree polynomial 𝑃 ( 𝑥 ) = 3 𝑥 4 + 5 𝑥 3 + 4 𝑥 2 + 3 𝑥 + 1 P(x)=3x 4 +5x 3 +4x 2 +3x+1. The polynomial can be rewritten as: 𝑃 ( 𝑥 ) = ( ( 3 𝑥 + 5 ) 𝑥 + 4 ) 𝑥 + 3 ) 𝑥 + 1. P(x)=((3x+5)x+4)x+3)x+1. Using Horner's Method to evaluate 𝑃 ( 2 ) P(2): 𝑃 ( 2 ) = ( ( ( 3 ( 2 ) + 5 ) 2 + 4 ) 2 + 3 ) 2 + 1 P(2)=(((3(2)+5)2+4)2+3)2+1 = ( ( ( 6 + 5 ) 2 + 4 ) 2 + 3 ) 2 + 1 =(((6+5)2+4)2+3)2+1 = ( ( 11 ⋅ 2 + 4 ) ⋅ 2 + 3 ) ⋅ 2 + 1 =((11⋅2+4)⋅2+3)⋅2+1 = ( ( 22 + 4 ) ⋅ 2 + 3 ) ⋅ 2 + 1 =((22+4)⋅2+3)⋅2+1 = ( 26 ⋅ 2 + 3 ) ⋅ 2 + 1 =(26⋅2+3)⋅2+1 = ( 52 + 3 ) ⋅ 2 + 1 =(52+3)⋅2+1 = 55 ⋅ 2 + 1 = 110 + 1 = 236. =55⋅2+1=110+1=236. Thus, 𝑃 ( 2 ) = 236 P(2)=236. (b) Horner's Method is the most efficient way of evaluating polynomials, as it minimizes the number of arithmetic operations. Question 3 (a) To calculate: How to rewrite any polynomial without powers of 𝑥 x. (b) To calculate: The relation between the new form and synthetic division. Answer: (a) A polynomial is written without powers of 𝑥 x by factoring 𝑥 x from the non-constant terms. Given 𝑃 ( 𝑥 ) = 3 𝑥 4 + 5 𝑥 3 + 4 𝑥 2 + 3 𝑥 + 1 P(x)=3x 4 +5x 3 +4x 2 +3x+1, it can be rewritten as: 𝑃 ( 𝑥 ) = ( ( ( 3 𝑥 + 5 ) 𝑥 + 4 ) 𝑥 + 3 ) 𝑥 + 1. P(x)=(((3x+5)x+4)x+3)x+1. This form uses addition and multiplication only. (b) The new form uses addition and multiplication in the same manner as synthetic division. Synthetic division evaluates the polynomial more efficiently. Question 4 To find: The real zeros of the polynomial function 𝑓 ( 𝑥 ) = 𝑥 4 − 4 𝑥 3 − 4 𝑥 2 + 16 𝑥 f(x)=x 4 −4x 3 −4x 2 +16x. Answer: The polynomial can be rewritten as: 𝑓 ( 𝑥 ) = 𝑥 ( 𝑥 3 − 4 𝑥 2 − 4 𝑥 + 16 ) . f(x)=x(x 3 −4x 2 −4x+16). Factoring: 𝑓 ( 𝑥 ) = 𝑥 ( 𝑥 + 2 ) 2 ( 𝑥 − 4 ) . f(x)=x(x+2) 2 (x−4). Real zeros: 𝑥 = − 2 , 0 , 2 , 4. x=−2,0,2,4. Question 5 To calculate: Polynomial equation with real coefficients that has roots 1 , 2 − 3 𝑖 1,2−3i. Answer: Formula Used: ( 𝑎 + 𝑏 ) ( 𝑎 − 𝑏 ) = 𝑎 2 − 𝑏 2 (a+b)(a−b)=a 2 −b 2 . If a polynomial has real coefficients, imaginary roots occur in conjugate pairs. Given roots: 1 , 2 − 3 𝑖 , 2 + 3 𝑖 1,2−3i,2+3i. Polynomial equation: ( 𝑥 − 1 ) [ ( 𝑥 − ( 2 − 3 𝑖 ) ) ( 𝑥 − ( 2 + 3 𝑖 ) ) ] = 0. (x−1)[(x−(2−3i))(x−(2+3i))]=0. ( 𝑥 − 1 ) [ ( 𝑥 − 2 ) 2 − ( 3 𝑖 ) 2 ] = 0. (x−1)[(x−2) 2 −(3i) 2 ]=0. ( 𝑥 − 1 ) [ ( 𝑥 − 2 ) 2 + 9 ] = 0. (x−1)[(x−2) 2 +9]=0. Expand: ( 𝑥 − 1 ) ( 𝑥 2 − 4 𝑥 + 13 ) = 0. (x−1)(x 2 −4x+13)=0. Final polynomial: 𝑥 3 − 5 𝑥 2 + 12 𝑥 − 13 = 0. x 3 −5x 2 +12x−13=0.