Questions and Answers #11 Polynomial Arithmetic

Questions and Answers Sheet 11 Polynomial Arithmetic Question #1 How can I prove that 4𝑛 + 15𝑛 − 1 ≡ 0(𝑚𝑜𝑑 9) ? Answer: By mathematical induction. Clearly, the statement is true for n=1 . Let it be valid for some natural number n. Is it then valid for n+1 ? Let us look at the difference (4𝑛+1 + 15(𝑛 + 1) − 1) − (4𝑛 + 15𝑛 − 1) = 3(4𝑛 ) + 15. We see that the statement for n+1 is valid if 3(4𝑛 ) + 15 is divisible by 9. But is it? Let us check it again using mathematical induction. For n=1 , the number 3(4𝑛 ) + 15 is clearly divisible by 9. So let us assume that 3(4𝑛 ) + 15 is divisible by 9 for some n, and let us look at 3(4𝑛+1 ) + 15. We see that the difference (3(4𝑛+1 ) + 15) − (3(4𝑛 ) + 15) = 9(4𝑛 ), is divisible by 9, which means that 3(4𝑛+1 ) + 15 is divisible by 9, which means that 4𝑛+1 + 15(𝑛 + 1) − 1 is divisible by 9. Question #2 What is the remainder when the polynomial 𝑥 + 𝑥 3 + 𝑥 9 + 𝑥 27 + 𝑥 81 + 𝑥 243 is divided by 𝑥 2 − 1 ? Answer: Since (𝑥 2 − 1) divides (𝑥 2𝑛 − 1) for each 𝑛 ∈ 𝑁 , 𝑥 2𝑛+1 = 𝑥(𝑥 2𝑛 − 1) + 𝑥 shows that 𝑥 2𝑛+1 leaves a remainder x when divided by 𝑥 2 − 1 whenever 𝑛 ∈ 𝑁 . Therefore, a sum of k odd powers of x leaves a remainder kx when divided by 𝑥 2 − 1. For 𝑥 + 𝑥 3 + 𝑥 9 + 𝑥 27 + 𝑥 81 + 𝑥 243 , 𝑘 = 6 and the remainder is 6x. Question #3 If 𝐷 = 1 − 𝑥 2 and 𝐶 = −1 − 3𝑥 2 + 4𝑥, find an expression that equals 3D-3C in standard form. Answer: 𝐷 = 1 − 𝑥 2 , 𝐶 = −1 − 3𝑥 2 + 4𝑥 3D-3C=3(D-C) = 3[1 − 𝑥 2 + 1 + 3𝑥 2 − 4𝑥) 3𝐷 − 3𝐶 = 3[2𝑥 2 − 4𝑥 + 2] 3𝐷 − 3𝐶 = 6𝑥 2 − 12𝑥 + 6 or ⇒ 6𝑥 2 − 6𝑥 − 6𝑥 + 6 ⇒ 6𝑥(𝑥 − 1) − 6(𝑥 − 1) =(x-1)(6x-6) =6(x-1)(x-1) 3𝐷 − 3𝐶 = 6(𝑥 − 1)2 Question #4 Building on standards from Analytical Geometry and from middle school, Students will analyze quadratic functions. Students will investigate key features of graphs, solve quadratic equations by taking the square roots, factoring (𝑥 2 + 𝑏𝑥 + 𝑐 and 𝑎𝑥 2 + 𝑏𝑥 + 𝑐), completing the square, and using the quadratic formula. Students will compare graphs in standard, vertex, andintercept forms. Students will only work with real number solutions. Answer: Quadratic eq-n is of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 where 𝑎 ≠ 0 b, c and c arbitrary constant. Given: 𝑥 2 + 𝑏𝑥 + 𝑐 = 0 𝑏 𝑏 2 𝑏 2 2 2 𝑏2 2 ⇒ 𝑥2 + 2 ⋅ 𝑥 ⋅ + ( ) − ( ) + 𝑐 = 0 𝑏 2 ⇒ (𝑥 + ) + (𝑐 − 2 4 )=0 2 𝑏 2 𝑏2 ⇒ (𝑥 + ) − (√( − 𝑐)) = 0. 2 4 𝑏 𝑏2 2 4 ⇒ (𝑥 + + √ 6 𝑏2 2 4 − 𝑐) (𝑥 + − √ 𝑏 𝑏2 2 4 ⇒ 𝑥 = −( +√ − 𝑐) = 0 − 𝑐) 𝑏 𝑏2 2 4 and 𝑥 = − ( − √ − 𝑐) 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 𝑏 𝑐 𝑎 𝑎 ⇒ 𝑎 (𝑥 2 + 𝑥 + ) = 0(𝑎 ≠ 0) ⇒ 𝑥2 + 2 ⋅ 𝑥 ⋅ 𝑏 𝑏 2𝑎 2 2 𝑏 𝑐 +( ) −( ) + =0 2𝑎 2𝑎 𝑎 2  (x + ⇒ (𝑥 + ⇒ (𝑥 + b c b2 + − 2 =0 2a a 4a 𝑏 2𝑎 2 ) − (√ 𝑏 2𝑎 ⇒ 𝑥 = −( +√ 𝑏 2𝑎 and 𝑥 = − ( 𝑏2 4𝑎2 +√ 𝑐 4𝑎 𝑎 𝑏 2 − ) = 0 𝑐 𝑏 𝑎 2𝑎 − ) (𝑥 + 𝑏2 −√ 𝑏2 4𝑎2 𝑐 − )=0 𝑎 𝑐 4𝑎2 2𝑎 2 𝑏2 −√ − ) 𝑎 𝑏2 4𝑎2 𝑐 − ) 𝑎 Question #5 Find all excluded for the expression. That is, find all values of x for which the expression is undefined. 3 8−2𝑥 x=? . Answer: The given expression is 3 8−2𝑥 . We know division by zero is undefined. So denominater must be equal to 0. i.e. 8-2x= or 2x=8 Answer: x=4 Question #6 If 𝐵 = 𝑝 + 5 and 𝐴 = 𝑝2 + 10𝑝 + 1, find an expression that equals 2B+3A in standard form. Answer: Given: B=p+5 𝐴 = 𝑝2 + 10𝑝 + 1 Here, the objective is to find the value of the expression 2B+3A. To find the value of the expression 2B+3A Substitute 𝐵 = 𝑝 + 5 and 𝐴 = 𝑝2 + 10𝑝 + 1 =2B+3A = 2(𝑝 + 5) + 3(𝑝2 + 10𝑝 + 1) = 2𝑝 + 10 + 3𝑝2 + 30𝑝 + 3 3𝑝2 + 32𝑝 + 13 Hence, the value of the expression 2𝐵 + 3𝐴 is 3𝑝2 + 32𝑝 + 13. Question #7 Your answer should be an expanded polynomial in standard form. (−3𝑦 2 − 5𝑦 − 2) + (−7𝑦 2 + 5𝑦 + 2) =? Answer: (−3𝑦 2 − 5𝑦 − 2) + (−7𝑦 2 + 5𝑦 + 2) = −3𝑦 2 − 5𝑦 − 2 − 7𝑦 2 + 5𝑦 + 2 = (−3𝑦 2 − 7𝑦 2 ) − 5𝑦 + 5𝑦 − 2 + 2 = (−10𝑦 2 ) + 0 + 0 = −10𝑦 2 −10𝑦 2 is the polynomial in standard form. Question #8 In this problem we are working over the field 𝑍7 , the integers mod 7. All arithmetic is done mod 7. First, write out the addition and multiplication tables for 𝑍7 . Secondly, we want codewords to be sequences of elements from 𝑍7 . Suppose that in order to detect and correct a possible error we will add two additional digits to the sequence. In this example, we have a sequence of length 3 that we wish to transmit, call it y1, y2, y3, so we find a polynomial f of degree at most 2 for which 𝑓(1) = 𝑦1, 𝑓(2) = 𝑦2, 𝑓(3) = 𝑦3 (all of this is done over 𝑍7 .) (To find f we use the 6 functions I discussed in class, but again the arithmetic is done mod 7.) To detect and correct a possible error we compute 𝑦4 = 𝑓(4) and 𝑦5 = 𝑓(5) and then transmit the sequence y1, y2, y3, y4, y5. If the received sequence is 1,4, 0, 6, 0 and there is at most one error, determine what the correct sequence is. (Again, remember that you are working mod 7- that will simplify the arithmetic slightly.) Answer: Consider the field 𝑍𝟟 = {0,1,2,3,4,5,6}𝑚𝑜𝑑 7 Solution: Consider the field 𝑍𝟟 = {0,1,2,3,4,5,6}𝑚𝑜𝑑7 We first write the addition and multiplication tables for 𝑍𝟟 + 0 1 2 (3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2) 3 4 5 The multiplicative Tables for 𝑍𝟟  0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 1 3 5 3 0 3 6 2 5 1 4 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 2 6 0 6 5 4 3 2 1 Question #9 Given the recursive Formula, write the explicit formula 1) 𝑎1 = 5 𝑎𝑛 = 𝑎𝑛−1 + 3 Given the explicit formula, write the recursive formula 2) 𝑎𝑛 = 15 + 4𝑛 Answer: 1) 𝑎1 = 5 𝑎𝑛 = 𝑎𝑛−1 + 3 ∴ 𝑎𝑛 − 𝑎𝑛−1 = 3 ∴ 𝑑 = 𝑎𝑛 − 𝑎𝑛−1 ∴𝑑=3 Therefore, 𝑎𝑛 = 𝑎1 + (𝑛 − 1)𝑑 𝑎𝑛 = 5 + (𝑛 − 1)3 𝑎𝑛 = 2 + 3𝑛 2) 𝑎𝑛 = 15 + 4𝑛 (1) 𝑎𝑛 = 𝑎 + (𝑛 − 1)𝑑 𝑎𝑛 = (𝑎 − 𝑑) + 𝑛𝑑 (2) ∴𝑑=4 𝑎1 − 𝑑 = 15 𝑎1 = 19 ∴ 𝑑 = 𝑎𝑛 − 𝑎𝑛−1 ∴ 𝑎𝑛 − 𝑎𝑛−1 = 4 ⇒ 𝑎𝑛 = 𝑎𝑛−1 + 4 Question #10 Determine whether the sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, give the common difference, if geometric, give the common ratio. − 1 1 1 1 , − , − , − , …. 42 35 28 21 Answer: 𝑎1 = − 1 41 , 𝑎2 = − 𝑎2 − 𝑎1 = −6 1435 1 35 , 𝑎3 = − , 𝑎3 − 𝑎2 = − 140 𝑎2 − 𝑎1 ≠ 𝑎3 − 𝑎2 Not Arithmetic Progression 𝑎2 6 𝑎 5 = , 3= 𝑎1 𝑎2 𝑎1 ≠ 𝑠 𝑎2 𝑎3 1 28 1 4 𝑎2 Not Geometric Progression