Explanation Procedure used: Factoring general trinomials by grouping : In a trinomial of the form a x 2 + b x + c , a ≠ 1 , split up the x -term into two terms using the method described below. “Step 1: Find the product a c . Step 2: Find two numbers whose product is a c and whose sums is b . Step 3: Replace the middle term bx with two x -terms having these numbers as coefficients. That is, a x 2 + b x + c = a x 2 + m x + n x + c . Step 4: Complete the factorization by grouping”. Calculation: Factor the given expression 6 x 2 + x − 5 as follows. It is observed that, the coefficients are a = 6 , b = 1 and c = − 5 . Step 1: Find the value of a c . Obtain the value of a c as, a c = 6 ( − 5 ) = − 30 Thus, the value of a c is –30. Step 2: Find the two number, m and n, whose product is a c = − 30 and whose sum is b = 1 . Obtain the list of integer factors of − 30 such that product is − 30 and sum is 1 as shown below in Table 1. Integers whose product is − 30 1 , − 30 15 , − 2 − 15 , 2 6 , − 5 − 6 , 5 Sum –29 13 –13 1 –1 Table 1 From Table 1, it is observed that the numbers whose product is − 30 and sum is 1 are − 5 and 6

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 6.2, Problem 31E

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To factor: The given expression 6x2+x−5 completely.

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Q1) Classify the following statements as a true or false statements a. Any ring with identity is a finitely generated right R module.- b. An ideal 22 is small ideal in Z c. A nontrivial direct summand of a module cannot be large or small submodule d. The sum of a finite family of small submodules of a module M is small in M A module M 0 is called directly indecomposable if and only if 0 and M are the only direct summands of M f. A monomorphism a: M-N is said to split if and only if Ker(a) is a direct- summand in M & Z₂ contains no minimal submodules h. Qz is a finitely generated module i. Every divisible Z-module is injective j. Every free module is a projective module Q4) Give an example and explain your claim in each case a) A module M which has two composition senes 7 b) A free subset of a modale c) A free module 24 d) A module contains a direct summand submodule 7, e) A short exact sequence of modules 74.

************* ********************************* Q.1) Classify the following statements as a true or false statements: a. If M is a module, then every proper submodule of M is contained in a maximal submodule of M. b. The sum of a finite family of small submodules of a module M is small in M. c. Zz is directly indecomposable. d. An epimorphism a: M→ N is called solit iff Ker(a) is a direct summand in M. e. The Z-module has two composition series. Z 6Z f. Zz does not have a composition series. g. Any finitely generated module is a free module. h. If O→A MW→ 0 is short exact sequence then f is epimorphism. i. If f is a homomorphism then f-1 is also a homomorphism. Maximal C≤A if and only if is simple. Sup Q.4) Give an example and explain your claim in each case: Monomorphism not split. b) A finite free module. c) Semisimple module. d) A small submodule A of a module N and a homomorphism op: MN, but (A) is not small in M.

Prove that Σ prime p≤x p=3 (mod 10) 1 Ρ = for some constant A. log log x + A+O 1 log x "

Chapter 6 Solutions

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

Ch. 6.1 - Prob. 1PE Ch. 6.1 - Prob. 2PE Ch. 6.1 - Prob. 3PE Ch. 6.1 - Prob. 4PE Ch. 6.1 - Prob. 1E Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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To factor: The given expression 6 x 2 + x − 5 completely.

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