
Polynomials (Chapter 4) - Core 1 Revision 1. The polynomial p(x
... The polynomials f(x) and g(x) are defined by f(x) = x3 + px2 – x + 5 g(x) = x3 – x2 + px +1 where p is a constant. When f(x) and g(x) are divided by x – 2, the remainder is R in each case. Find the values of p and R (Total 5 marks) ...
... The polynomials f(x) and g(x) are defined by f(x) = x3 + px2 – x + 5 g(x) = x3 – x2 + px +1 where p is a constant. When f(x) and g(x) are divided by x – 2, the remainder is R in each case. Find the values of p and R (Total 5 marks) ...
Cyclic groups and elementary number theory
... As before, 0 ≤ r2 − r1 ≤ r2 ≤ n − 1 < n. Since n is the order of g, and hence the smallest positive power k of g such that g k = 1, we must have r2 − r1 = 0, i.e. r1 = r2 . This proves the uniqueness part of (i). (ii) Let n ∈ Z be arbitrary. Then N = nq + r, where 0 ≤ r ≤ n − 1, and n|N ⇐⇒ r = 0. T ...
... As before, 0 ≤ r2 − r1 ≤ r2 ≤ n − 1 < n. Since n is the order of g, and hence the smallest positive power k of g such that g k = 1, we must have r2 − r1 = 0, i.e. r1 = r2 . This proves the uniqueness part of (i). (ii) Let n ∈ Z be arbitrary. Then N = nq + r, where 0 ≤ r ≤ n − 1, and n|N ⇐⇒ r = 0. T ...
N.4 - DPS ARE
... ELG.MA.HS.N.4: Perform arithmetic operations with complex numbers. N-CN.A.1 Know there is a complex number i such that i2 = –1, and every complex number has the form a + bi with a and b real. N-CN.A.2 Use the relation i2 = –1 and the commutative, associative, and distributive properties to add, ...
... ELG.MA.HS.N.4: Perform arithmetic operations with complex numbers. N-CN.A.1 Know there is a complex number i such that i2 = –1, and every complex number has the form a + bi with a and b real. N-CN.A.2 Use the relation i2 = –1 and the commutative, associative, and distributive properties to add, ...
Unit 4 Lesson 1 Day 5
... of 0, and each substitution reduces the degree on the part of the polynomial that needs to be divided into linear factors. After these two substitutions, we can rewrite the polynomial as a product of two linear and one quadratic factor: (x – 2i)(x + 2i)(x2 – x – 1) We can use the quadratic formula t ...
... of 0, and each substitution reduces the degree on the part of the polynomial that needs to be divided into linear factors. After these two substitutions, we can rewrite the polynomial as a product of two linear and one quadratic factor: (x – 2i)(x + 2i)(x2 – x – 1) We can use the quadratic formula t ...
Homework 10 April 13, 2006 Math 522 Direction: This homework is
... construct a table that convert polynomials in F # to powers of z, and vice versa. Here F # means the nonzero elements of the field F . Answer: The conversion table can be constructed using the following maple commands: > f := x− > x4 + x + 1: > z := x2 + 1: > for i from 1 to 15 do > temp := Powmod(z ...
... construct a table that convert polynomials in F # to powers of z, and vice versa. Here F # means the nonzero elements of the field F . Answer: The conversion table can be constructed using the following maple commands: > f := x− > x4 + x + 1: > z := x2 + 1: > for i from 1 to 15 do > temp := Powmod(z ...
SOME PARI COMMANDS IN ALGEBRAIC NUMBER
... factor(n) factors the integer n into primes. (This works on rational numbers also and will give prime factorizations with negative exponents.) gcd(a,b) is the greatest common divisor of a and b. isprime(n) returns 1 if n is prime and 0 otherwise. prime(n) returns the nth prime. primes(n) is a vector ...
... factor(n) factors the integer n into primes. (This works on rational numbers also and will give prime factorizations with negative exponents.) gcd(a,b) is the greatest common divisor of a and b. isprime(n) returns 1 if n is prime and 0 otherwise. prime(n) returns the nth prime. primes(n) is a vector ...
Monte Carlo Methods in Scientific Computing
... crystal is drastically softened in the porous glass and becomes continuous, an effect that was not attributed to finitesize effects but rather to the influence of ...
... crystal is drastically softened in the porous glass and becomes continuous, an effect that was not attributed to finitesize effects but rather to the influence of ...
Math 5c Problems
... A3. Galois theory thus tells us that L Q[] where is any root of g(x). Two of the roots are = 8 cos( /9) and 0 = 8 cos(5 /9). Find polynomials p; q such that p() = 0 and q() = 00 where 00 is the third root. b) Let 1 + 2 + 3 + 4 be the roots of f (x) and let H1 = f1; (123); (132 ...
... A3. Galois theory thus tells us that L Q[] where is any root of g(x). Two of the roots are = 8 cos( /9) and 0 = 8 cos(5 /9). Find polynomials p; q such that p() = 0 and q() = 00 where 00 is the third root. b) Let 1 + 2 + 3 + 4 be the roots of f (x) and let H1 = f1; (123); (132 ...
Jumping Jiving GCD - the School of Mathematics, Applied
... Exercise: In a country all money are of two types: Golden coins worth 115 EU each and Silver coins worth 45 EU each. a) Which prices can they use in this country and why? b) You want to buy an item costing 5 EU. What are all the ways this transaction be made? c) How about for an item costing 30 EU? ...
... Exercise: In a country all money are of two types: Golden coins worth 115 EU each and Silver coins worth 45 EU each. a) Which prices can they use in this country and why? b) You want to buy an item costing 5 EU. What are all the ways this transaction be made? c) How about for an item costing 30 EU? ...
Math 307 Abstract Algebra Homework 7 Sample solution Based on
... (b) What is the order of 4U5 (105) in the factor group U (105)/U5 (105). Solution. We must find the smallest m such that 4m ∈ U5 (105), so 4m must be relatively prime to 105 and 4m = 5k + 1, 0 ≤ k ≤ 20. 105 = 3 ∗ 5 ∗ 7, and 4m = 22m , so 4m will always be relatively prime to 105 because they share n ...
... (b) What is the order of 4U5 (105) in the factor group U (105)/U5 (105). Solution. We must find the smallest m such that 4m ∈ U5 (105), so 4m must be relatively prime to 105 and 4m = 5k + 1, 0 ≤ k ≤ 20. 105 = 3 ∗ 5 ∗ 7, and 4m = 22m , so 4m will always be relatively prime to 105 because they share n ...
Constructions of plane curves with many points
... interesting examples this way. 4. Integral points on the curve Pd (x, y) = 0. In [5], certain polynomials Pd (x, y) ∈ Z[x, y] of degree d were constructed and it was shown that Pd for every d is absolutely irreducible and has at least d2 + 2d + 3 integral solutions to Pd (x, y) = 0. In his review [2 ...
... interesting examples this way. 4. Integral points on the curve Pd (x, y) = 0. In [5], certain polynomials Pd (x, y) ∈ Z[x, y] of degree d were constructed and it was shown that Pd for every d is absolutely irreducible and has at least d2 + 2d + 3 integral solutions to Pd (x, y) = 0. In his review [2 ...
Lecture Thursday
... positive and negative x we see that f (x) > 0 if and only if x > 0. There is a much simpler way to get to this result which avoids any algebraic manipulation of inequalities: f (x) is the product of the two factors x2 + 1 and x. In order for f (x) to be positive either both factors must be positive ...
... positive and negative x we see that f (x) > 0 if and only if x > 0. There is a much simpler way to get to this result which avoids any algebraic manipulation of inequalities: f (x) is the product of the two factors x2 + 1 and x. In order for f (x) to be positive either both factors must be positive ...