Questions and Answers #6 Polynomial Arithmetic

Questions and Answers Sheet 6 Polynomial Arithmetic Question #1 The solution to fill the blank in the statement so that the resulting statement is true. The statement in which we have to fill the blank is ''The degree of the monomial −18𝑥 4 𝑦 2 is ___________________'' Answer: Given: Fill in the blank of the statement ''The degree of the monomial −18𝑥 4 𝑦 2 is ____'' so that the resulting statement is true. Algebraic expressions are created by combining numbers and variables using arithmetic operations such as addition, subtraction, multiplication, division, and exponentiation. Using these mathematical operations, we can create an expression called a polynomial. A polynomial is a sum of or difference of terms, each consisting of a variable rose to a nonnegative integer power. A polynomial which contains only one term is called Monomial, polynomial which contains two terms is called Binomial, and a polynomial which contains three terms is called Trinomial. A polynomial is a sum of monomials where each monomial is called a term. It is usually written with the term with the highest exponent of the variable first and then decreasing from left to right. The first term of a polynomial is called the leading coefficient. The number term of a polynomial which is not multiplied by a variable is called a constant. The basic building block of a polynomial is a monomial. A monomial is one term which can be a number, a variable, or the product of a number and variables with an exponent. The number part of the term is called the coefficient. The coefficient can be any real number, including 0. Exponent of the variable must be a whole number. We should keep in mind that a monomial cannot have a variable in the denominator or a negative exponent. The value of the exponent is the degree of a monomial. We should also consider that a variable that appears to have no exponent really has an exponent of 1 and a monomial with no variable has a degree of 0. In a monomial of the form 𝑎𝑥 𝑚 𝑦 𝑛 , a is the coefficient of the monomial and m+n be the degree of the monomial. There by the given monomial −18𝑥 4 𝑦 2 , the degree of monomial is 6(4+2) Conclusion: The degree of the monomial −18𝑥 4 𝑦 2 is 6. Question #2 Write the polynomial in standard form. Then classify the polynomial by degree and by a number of terms. 2𝑥 5 + 8𝑥 5 − 𝑥 5 Write the polynomial in standard form (simplify your answer) Answer: Step 1 To write a polynomial in standard form it must be from largest to the smallest exponent. Degree of a polynomial is the highest exponent in the polynomial. Number of terms in the polynomial means how many terms are there in the polynomial. Step 2 The given polynomial is, 2𝑥 5 + 8𝑥 5 − 𝑥 5 = 10𝑥 5 − 𝑥 5 = 9𝑥 5 2𝑥 5 + 8𝑥 5 − 𝑥 5 = 9𝑥 5 Standard form is 9𝑥 5 Degree fo the polynomial is 5. Number of rerms in the polynomial is 1. Question #3 To find: The sum of the given arithmetic sequence. Given: the sequence ∑ 𝑛 = 190(3 − 2𝑛) Answer: Formula used: The formula for the summation of a polynomial with degree 1 is, ∑ 𝑘 = 1𝑛𝑘 = 𝑛2(𝑛 + 1) … . (1) The formula for the summation of a constant is, ∑ 𝑘 = 1𝑛𝑐 = 𝑐𝑛 … . . (2) Calculation: To find the sum of the arithmetic sequence, i.e, to find, 𝑆𝑛 = ∑ 𝑛 = 190(3 − 2𝑛) Write ∑ 𝑛 = 1903 − 2𝑛 as follows, ∑ 𝑛 = 190(3 − 2𝑛) = ∑ 𝑛 = 1903 − 2 ∑ 𝑛 = 190𝑛 Now to find ∑ 𝑛 = 1903, Substitute the values of n in equation (2) We get, ∑ 𝑛 = 1903 =3(90) =270 Now to find the 2 ∑ 𝑛 = 190𝑛 Substitute the values of n in equation (1). We get, 2 ∑ 𝑛 = 190𝑛 = 2(90(90 + 1)2) = 2(90 × 912) =8190 Then, ∑ 𝑛 = 1903 + 2 ∑ 𝑛 = 190𝑛 = 270 − 8190 =-7920 Hence ∑ 𝑛 = 190(3 − 2𝑛) = −7920. Question #4 Arithmetic or Geometric? a) If 𝑎1 , 𝑎2 , 𝑎3 , ⋯ is an arithmetic sequence, is the sequence 𝑎1 + 2, 𝑎2 + 2, 𝑎3 + 2, ⋯ arithmetic? b) If 𝑎1 , 𝑎2 , 𝑎3 , ⋯ is a geometric sequence, is the sequence 5𝑎1 , 5𝑎2 , 5𝑎3 , ⋯ geometric? Answer: a) To show: the sequence 𝑎1 + 2, 𝑎2 + 2, 𝑎3 + 2, ⋯ is arithmetic. Given: The sequence 𝑎1 , 𝑎2 , 𝑎3 , ⋯ is an arithmetic sequence. Approach: In arithmetic sequence, difference between the terms remains constant throughout. Calculation: The sequence 𝑎1 , 𝑎2 , 𝑎3 , ⋯ is an arithmetic sequence. 𝑎2 − 𝑎1 = 𝑎3 − 𝑎2 = ⋯ ⋯ (1) (𝑎2 + 2) − (𝑎1 + 2) = (𝑎2 − 𝑎1 ) (𝑎3 + 2) − (𝑎2 + 2) = (𝑎3 − 𝑎2 ) From equation (1) (𝑎3 + 2) − (𝑎2 + 2) = (𝑎2 − 𝑎1 ) As the differences between the terms are same then it is an arithmetic sequence. Conclusion: hence, the sequence 𝑎1 + 2, 𝑎2 + 2, 𝑎3 + 2, ⋯ is arithmetic. b) To show: 5𝑎1 , 5𝑎2 , 5𝑎3 , ⋯ is a geometric sequence. Given: The sequence 𝑎1 , 𝑎2 , 𝑎3 , ⋯ is geometric sequence. Approach: In a geometric sequence each term is found by multiplying the previous term by a constant. Calculation: 𝑎2 𝑎 = 3 = ⋯ ⋯ (1) 𝑎1 𝑎2 Multiplying equation (1) by 5 5𝑎2 5𝑎 = 3=⋯ 5𝑎1 5𝑎2 Then it is a geometric sequence. Conclusion: hence, the sequence 5𝑎1 , 5𝑎2 , 5𝑎3 , ⋯ is a geometric sequence. Question #5 Determine the number of terms there are in the finite arithmetic sequence. 3, −2, −7, ⋯ , −87 Answer: a= first term d= common difference l= last form here given that a=3 d=-2-3=-5 l=87 we also know that l=a+(n-1)d -87=3+(n-1)(-5) =87=3-5n+5 -87=8-5n 5n=8+87 5n=95 n=19 Here are 19 term in given arithmetic sequence. Question #6 Find the number of terms in the arithmetic sequence with the given conditions. 29 1 𝑎1 = − , 𝑑 = − , 𝑆 = −36 6 3 Answer: Step 1 The first term of the arithmetic sequence is given as − The common difference of the arithmetic sequence is 29 1 6 3 The sum of the terms is given as -36 Step 2 If the number of terms is n, then the sum of the arithmetic sequence is given as 𝑛 𝑆𝑛 = [2𝑎1 + (𝑛 − 1)𝑑] 2 𝑛 29 2 6 1 −36 = [2 ⋅ (− ) + (𝑛 − 1) ⋅ ] −72 = 𝑛 [− −72 = − −72 = 29 29𝑛 6 + (𝑛 − 1) ] + 𝑛(𝑛 − 1) 3 −29𝑛+𝑛(𝑛−1) 3 3 1 3 1 3 −72 = −29𝑛+𝑛2 −𝑛 3 −216 = −30𝑛 + 𝑛2 𝑛2 − 30𝑛 + 216 = 0 Using the quadratic equation formula, we get 𝑛= = = = = = −(−30)±√(−30)2 −4(1)(216) 2⋅1 30±√900−864 2 30±√36 2 30±6 2 30−6 2 24 2 30+6 and and 2 36 2 =12 and 18 Therefore, the number of terms is either 12 or 18. Question #7 The scoring of modern IQ tests is such that Intelligence Quotients (IQs) have Normal distribution with 𝜇 = 100 and 𝜎 = 15 a) What percent of people have IQ less than 80? b) What percent of people have IQ greater than 120? c) Mensa International is a non-profit organization that accepts only people with IQ within the top 2%. What level of IQ qualifies one to be a member of Mensa? Answer: Step 1 Given: 𝜇 = 𝑀𝑒𝑎𝑛 = 100 𝜎 = Standard deviation = 15 a) The z-score is the value decreased by the mean, divided by the standard deviation. 𝑥−𝜇 80−100 𝑧= = ≈ −1.33 𝜎 15 Determine the corresponding probability using the normal probability table in the appendix. 𝑃(𝑍 < −1.33) is given in the row starting with -1.3 and in the column starting with 0.03 of the standard normal probability table in the appendix. 𝑃(𝑋 < 80) = 𝑃(𝑍 < −1,33) =0.0918 =9.18% Thus 9.18% of the people are expected to have an IQ less than 80. Step 2 b) The z-score is the value decreased by the mean, divided by the standard deviation. 𝑥−𝜇 120−100 𝑧= = ≈ 1.33 𝜎 15 Determine the corresponding probability using the normal probability table in the appendix. 𝑃(𝑍 < 1.33) is given in the row starting with 1.3 and in the cilumn starting with 0.03 of the standard normal probability table in the appendix. 𝑃(𝑋 > 120) = 𝑃(𝑍 > 1,33) =1-P(Z<1.33) =1-0.9082 =9.18% Thus 9.18% of the people are expected to have an IQ less than 120. c) The boundary of the top 2% has the property that 100%-2%=98% of the people below this boundary. Determine the critical value corresponding with a probability of 98% or 0.98 using the normal probability table in the appendix. We note that the closest probability 0.9798 is given in the row starting with ''2.0'' and in the column starting with ''0.05: z=2.05 The corresponding value is then obtained by adding the product of the z-score and the standard deviation to the mean: 𝑥 = 𝜇 + 𝑧𝜎 = 100 + 2.05(15) = 130.75 Thus the top 2% have an IQ of at least 130.75. Question #8 The pathogen Phytophthora capsici causes bell pepper plants to wilt and die. A research project was designed to study the effect of soil water content and the spread of the disease in fields of bell peppers. It is thought that too much water helps spread the disease. The fields were divided into rows and quadrants. The soil water content (percent of water by volume of soil) was determined for each plot. An important first step in such a research project is to give a statistical description of the data. Soil Water Content for Bell Pepper Study 15 9 15 10 14 14 14 13 12 11 11 11 11 10 11 13 16 10 15 12 9 10 7 14 13 14 8 9 8 11 13 13 12 9 10 9 9 16 16 12 10 11 11 12 15 6 10 10 11 9 Make a frequency table using four classes. Then estimate the mean and sample standard deviation using the frequency table. Compute a 75% Chebyshev interval centered about the mean. Answer: Step 1 The class width is the difference between the largest and smallest value, divided by the number of classes (round up to the nearest integer!). 16−6 Class Wigth = = 2.5 ≈ 3 4 Determine the midpoints, frequencise, the products of the midpoints and the frequencies, and the products of the squared midpoints and the frequencies. Interval Midpoint x f xf x2 f 6−8 7 4 28 196 9 − 11 10 24 240 2400 12 − 14 13 15 195 2535 15 − 17 16 7 112 1792 SUM 50 575 6923 Step 2 The sample mean is then: 𝑥̅ = ∑ 𝑥𝑓 𝑛 = 575 50 = 11.5 The sample standard deviation is then: 𝑠=√ ∑ 𝑥 2 𝑓− (∑ 𝑥𝑓)2 𝑛 𝑛−1 =√ 5722 50 6923− 50−1 ≈ 2.5173 Step 3 Using Chebyshev's Rule with k=2, we know that at least 100 (1 − standard deviations from the mean. 1 𝑘2 1 ) % = 100 (1 − ) % = 75% is within 2 4 𝑥̅ − 2𝑠 = 11.5 − 2(2.5173) = 6.4654 𝑥̅ + 2𝑠 = 11.5 + 2(2.5173) = 16.5346 Question #9 Consider 𝐴(𝑥) = 𝑥 3 + 𝑥 2 + 1 and 𝐵(𝑥) = 𝑥 2 + 𝑥 + 1 in GF(11). Also, the irreducible polinomial is given as 𝑝(𝑥) = 𝑥 3 + 𝑥 + 1 1) Find A(x)+B(x) (10 pts) Hint: Use p(x) if the initial result does not fall in GF(11) 2. Find 𝐴2 (𝑥) ⋅ 𝐵(𝑥) (20 pts) Hint: Use p(x) if the initial does not fall in GF(11). Answer: 𝐴(𝑥) = 𝑥 3 + 𝑥 2 + 1 𝑏(𝑥) = 𝑥 2 + 𝑥 + 1 and 𝑝(𝑥) = 𝑥 3 + 𝑥 + 1 (a) find A(x)+b(x) 𝐴(𝑥) + 𝐵(𝑥) = (𝑋 3 + 𝑥2 + 1) + (𝑥 2 + 𝑥 + 1) = 𝑥 3 + 2𝑥 2 + 𝑥 + 2 𝐴(𝑥) + 𝐵(𝑥) = 𝑥 3 + 2𝑥 2 + 𝑥 + 2 = (𝑥 3 + 𝑥 + 1) + (2𝑥 2 + 1) 𝐴(𝑥) + 𝐵(𝑥) = 𝑝(𝑥) + (2𝑥 2 + 1) p(x) it is initial result. 𝐴(𝑥) + 𝐵(𝑥) = 𝑥 3 + 2𝑥 2 + 𝑥 + 2 (b). 𝐴2 (𝑥) ⋅ 𝐵(𝑥) =? 𝐴(𝑥) = (𝑥 3 + 𝑥 2 + 1)2 𝐴2 (𝑥) = (𝑥 3 + 𝑥 2 + 1) ⋅ (𝑥 3 + 𝑥 2 + 1) = 𝑥 6 + 𝑥 5 + 𝑥 3 + 𝑥 5 + 𝑥 4 + 𝑥 2 + 1 𝐴2 (𝑥) = 𝑥 6 + 2𝑥 5 + 2𝑥 3 + 𝑥 4 + 2𝑥 2 + 1 𝐴2 (𝑥) ⋅ 𝐵(𝑥) = (𝑥 6 + 2𝑥 5 + 2𝑥 3 + 𝑥 4 + 2𝑥 2 + 1) ⋅ (𝑥 2 + 𝑥 + 1) 𝐴2 (𝑥) ⋅ 𝐵(𝑥) = 𝑥 8 + 2𝑥 7 + 2𝑥 5 + 𝑥 6 + 2𝑥 4 + 𝑥 2 + 𝑥 7 + 2𝑥 6 + 2𝑥 4 + 𝑥 5 + 2𝑥 4 + 𝑥 + 𝑥 6 + 2𝑥 5 + 𝑥 3 + 𝑥 4 + 2𝑥 2 + 1 𝐴2 (𝑥) ⋅ 𝐵(𝑥) = 𝑥 8 + 37 + 4𝑥 6 + 5𝑥 5 + 5𝑥 4 + 4𝑥 3 + 3𝑥 2 + 𝑥 + 1 Question #10 This problem deals with polynomials in 𝑍2 [𝑋]. Remember these have 0-1 coefficients. You use the usual algorithms, but with all arithmetic done mod 2. a) It was observed in class that the constant term of an irreducible polynomial is nonzero. Show further that the Hamming weight (number of nonzero coeffs) of an irreducible polynomial of degree >>1 is odd. [Hint: Use the remainder theorem.] b) Only one degree 2 polynomial f meets the conditions of a). What is it? Give the addition and multiplication tables for polynomial arithmetic mod f. c) Using similar ideas, find all the irreducibles of degrees 3 and 4. (Note: For degree 4, you will need to consider a possible quadratic factor.) Answer: a) Suppose that any polynomial 𝑢𝑝 ⇒ 𝑝(𝑥) in 𝑍2 [𝑋] has even number of nonzero coefficient then p(1) is even (because the coefficient are eithes 0 or 1. So nonzero coefficient sum of coefficient which are even) and p(r) is even ⇒ 1 is root of p(x) with operation ⋅ 𝑚𝑜𝑑 𝑍 ⇒ If 1 is root of p(x) right ⇒ 𝑝(𝑥) is reducible. So, if a polynomial is irreducible with a constant term than the Hamming weight (No of nonzero coefficient) is add. (If constant term is not included then x=0 is a root and polynomial is reducible) b) We want a polynomial of degree Z with constant term which is irreducible i-e no of no of nonzero coefficients is add. Since polynomial of degree with constant term has at last two nonzero coefficient (one coefficient of 𝑥 2 and 2nd constant term). Se, for total coefficient to be add coefficient of x must be nonzero. So 𝑓(𝑥) = 𝑥 2 + 𝑥 + 1 = 𝑓(𝑥). Under mod f operation any polynomial has degree at most 1. So, polynomials under mod f are {1, 𝑥, 1 + 𝑥, 0} x 0 1 x 1+ x 0 1 x 1+ x 0 0 0 0 0 1 x 1+ x 0 x 1+ x 1 0 1+ x 1 x + 0 1 x 1+ x 0 0 1 x 1+ x 1 1 0 1+ x x x x 1+ x 0 1 1+ x 1+ x x 1 0 c) Irreducible polynomials of degree 3. No of nonzero coefficient ''with constant term'' are add in an ⇒ irreducible polynomial with constant term. So, For degree 3, irreducible polynomial 𝑥 4 + 𝑥 3 + 1, 𝑥 4 + 𝑥 2 + 1 (For a polynomial of degree with constant term 𝑥 3 and constants has coefficient 1: So remaining terms 𝑥 2 and x; For nonzero coefficient to be add we can add any one of 𝑥 2 and x) For polynomial of degree 4, polynomial with add no of nonzero coefficients and constant terms are 𝑥 4 + 𝑥 3 + 1, 𝑥 4 + 𝑥 2 + 1, 𝑥 4 + 𝑥 + 1, 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1 Now, if any polynomial f(x) of degree 4 can be factorized into 𝑓1 (𝑥) and 𝑓2 (𝑥) of degree 2, and f(x) has hamming no add. (If hamming no is even then f(x) is reducible.) Now f(x) has hamming no add. So 𝑓1 (𝑥) and 𝑓2 (𝑥) also has hamming no add. Since any of 𝑓1 (𝑥), 𝑓2 (𝑥) has hamming no even then 1 is root of 𝑓1 (𝑥) o r 𝑓2 (𝑥) ⇒ 1 is root of 𝑓1(𝑥) ⇒ 𝑓(𝑥) has even hamming no. So, 𝑓1 (𝑥) and 𝑓2 (𝑥) are 2 degree polynomial with add hamming no and by part (b). 𝑓1 (𝑥) = 𝑓2 (𝑥) = 𝑥 2 + 𝑥 + 1 So 𝑓(𝑥) = 𝑓2 (𝑥) = (𝑥 2 + 𝑥 + 1) + (𝑥 2 + 𝑥 + 1) = 𝑥 4 + 𝑥 2 + 1 So, 𝑥 4 + 𝑥 2 + 1 can be reduced into madratic forms. So, irreducible polynomial of degree 2 are 𝑥 4 + 𝑥 3 + 1, 𝑥 4 + 𝑥 + 1, 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1