Polynomials #1 - Questions and Answers

Question 1 Find a polynomial with integer coefficients that satisfies the given conditions. Q has degree 3 and zeros 0 and 𝑖 i. Answer: From the Complex Conjugate Zeros Theorem: If 𝑖 i is a zero of the polynomial 𝑄 ( 𝑥 ) Q(x) with real coefficients, its conjugate − 𝑖 −i is also a zero of 𝑄 ( 𝑥 ) Q(x). From the Factor Theorem: If 0 is a zero of 𝑄 ( 𝑥 ) Q(x), the expression ( 𝑥 − 0 ) (x−0) or 𝑥 x is a factor of 𝑄 ( 𝑥 ) Q(x). If 𝑖 i is a zero of 𝑄 ( 𝑥 ) Q(x), the expression ( 𝑥 − 𝑖 ) (x−i) is a factor of 𝑄 ( 𝑥 ) Q(x). If − 𝑖 −i is a zero of 𝑄 ( 𝑥 ) Q(x), the expression ( 𝑥 + 𝑖 ) (x+i) is a factor of 𝑄 ( 𝑥 ) Q(x). Thus, 𝑄 ( 𝑥 ) Q(x) is defined as: 𝑄 ( 𝑥 ) = 𝑥 ( 𝑥 − 𝑖 ) ( 𝑥 + 𝑖 ) . Q(x)=x(x−i)(x+i). Step 1: Simplify using the difference of squares: ( 𝑥 − 𝑖 ) ( 𝑥 + 𝑖 ) = 𝑥 2 − 𝑖 2 = 𝑥 2 + 1. (x−i)(x+i)=x 2 −i 2 =x 2 +1. Step 2: Substitute back: 𝑄 ( 𝑥 ) = 𝑥 ( 𝑥 2 + 1 ) = 𝑥 3 + 𝑥 . Q(x)=x(x 2 +1)=x 3 +x. Final Answer: 𝑄 ( 𝑥 ) = 𝑥 3 + 𝑥 . Q(x)=x 3 +x. Question 2 How do you factor 𝑥 3 − 8 x 3 −8? Answer: Using the difference of cubes formula: 𝑎 3 − 𝑏 3 = ( 𝑎 − 𝑏 ) ( 𝑎 2 + 𝑎 𝑏 + 𝑏 2 ) . a 3 −b 3 =(a−b)(a 2 +ab+b 2 ). For 𝑥 3 − 8 x 3 −8: 𝑥 3 − 8 = ( 𝑥 − 2 ) ( 𝑥 2 + 2 𝑥 + 4 ) . x 3 −8=(x−2)(x 2 +2x+4). Question 3 Form the polynomial in the usual way. Find the degree and leading coefficient of the polynomial. Next, organize the polynomial into groups based on the number of terms. Given: 6 𝑐 2 + 2 𝑐 − 𝑐 6c 2 +2c−c. Answer: Organize the polynomial: 6 𝑐 2 + 2 𝑐 − 𝑐 = 6 𝑐 2 + 𝑐 . 6c 2 +2c−c=6c 2 +c. Degree: 2 (highest power of 𝑐 c). Leading Coefficient: 6 (coefficient of 𝑐 2 c 2 ). Type: Binomial (2 terms). Question 4 Find the indicated products. Assume all variables that appear as exponents represent positive integers. ( 𝑥 2 𝑎 − 7 ) ( 𝑥 2 𝑎 − 3 ) (x 2 a−7)(x 2 a−3) Answer: Expand: ( 𝑥 2 𝑎 − 7 ) ( 𝑥 2 𝑎 − 3 ) = 𝑥 4 𝑎 − 3 𝑥 2 𝑎 − 7 𝑥 2 𝑎 + 21. (x 2 a−7)(x 2 a−3)=x 4a −3x 2a −7x 2a +21. Combine like terms: 𝑥 4 𝑎 − 10 𝑥 2 𝑎 + 21. x 4a −10x 2a +21. Final Answer: 𝑥 4 𝑎 − 10 𝑥 2 𝑎 + 21. x 4a −10x 2a +21. Question 5 Approximate 𝑓 ( 𝑥 ) f(x) by a Taylor polynomial with degree 𝑛 n at the number 𝑎 a: 𝑓 ( 𝑥 ) = ln ⁡ ( 1 + 2 𝑥 ) , 𝑎 = 1 , 𝑛 = 3 , 0.5 ≤ 𝑥 ≤ 1.5. f(x)=ln(1+2x),a=1,n=3,0.5≤x≤1.5. Answer: The Taylor polynomial with degree 𝑛 n at 𝑎 a is given by: 𝑓 ( 𝑥 ) ≈ 𝑇 𝑛 ( 𝑥 ) = ∑ 𝑘 = 0 𝑛 𝑓 ( 𝑘 ) ( 𝑎 ) 𝑘 ! ( 𝑥 − 𝑎 ) 𝑘 . f(x)≈T n ​ (x)= k=0 ∑ n ​ k! f (k) (a) ​ (x−a) k . Step 1: Compute derivatives of 𝑓 ( 𝑥 ) = ln ⁡ ( 1 + 2 𝑥 ) f(x)=ln(1+2x): 𝑓 ( 𝑥 ) = ln ⁡ ( 1 + 2 𝑥 )    ⟹    𝑓 ( 1 ) = ln ⁡ ( 3 ) f(x)=ln(1+2x)⟹f(1)=ln(3). 𝑓 ′ ( 𝑥 ) = 2 1 + 2 𝑥    ⟹    𝑓 ′ ( 1 ) = 2 3 f ′ (x)= 1+2x 2 ​ ⟹f ′ (1)= 3 2 ​ . 𝑓 ′ ′ ( 𝑥 ) = − 4 ( 1 + 2 𝑥 ) 2    ⟹    𝑓 ′ ′ ( 1 ) = − 4 9 f ′′ (x)= (1+2x) 2 −4 ​ ⟹f ′′ (1)= 9 −4 ​ . 𝑓 ′ ′ ′ ( 𝑥 ) = 16 ( 1 + 2 𝑥 ) 3    ⟹    𝑓 ′ ′ ′ ( 1 ) = 16 27 f ′′′ (x)= (1+2x) 3 16 ​ ⟹f ′′′ (1)= 27 16 ​ . Step 2: Taylor polynomial at 𝑎 = 1 a=1: 𝑇 3 ( 𝑥 ) = 𝑓 ( 1 ) + 𝑓 ′ ( 1 ) ( 𝑥 − 1 ) + 𝑓 ′ ′ ( 1 ) 2 ! ( 𝑥 − 1 ) 2 + 𝑓 ′ ′ ′ ( 1 ) 3 ! ( 𝑥 − 1 ) 3 . T 3 ​ (x)=f(1)+f ′ (1)(x−1)+ 2! f ′′ (1) ​ (x−1) 2 + 3! f ′′′ (1) ​ (x−1) 3 . Substitute values: 𝑇 3 ( 𝑥 ) = ln ⁡ ( 3 ) + 2 3 ( 𝑥 − 1 ) − 4 18 ( 𝑥 − 1 ) 2 + 16 162 ( 𝑥 − 1 ) 3 . T 3 ​ (x)=ln(3)+ 3 2 ​ (x−1)− 18 4 ​ (x−1) 2 + 162 16 ​ (x−1) 3 . Simplify: 𝑇 3 ( 𝑥 ) = ln ⁡ ( 3 ) + 2 3 ( 𝑥 − 1 ) − 2 9 ( 𝑥 − 1 ) 2 + 8 81 ( 𝑥 − 1 ) 3 . T 3 ​ (x)=ln(3)+ 3 2 ​ (x−1)− 9 2 ​ (x−1) 2 + 81 8 ​ (x−1) 3 . Final Answer: 𝑇 3 ( 𝑥 ) = ln ⁡ ( 3 ) + 2 3 ( 𝑥 − 1 ) − 2 9 ( 𝑥 − 1 ) 2 + 8 81 ( 𝑥 − 1 ) 3 . T 3 ​ (x)=ln(3)+ 3 2 ​ (x−1)− 9 2 ​ (x−1) 2 + 81 8 ​ (x−1) 3 .