Polynomial + binary arithmetic expert please help?
Q. From the text book: With the following polynomials: f(x) = x^6 + x^4 + x^2 + x + 1 g(x) = x^7 + x + 1 m(x) = x^8 + x^4 + x^3 + x + 1 f(x) * g(x) mod m(x) = x^7 + x^6 + 1 ___ To do it in binary arithmetic: Step1: (0101 0111) x (000 0001) = (1010 1110) Step2: (0101 0111) x (000 0100) = (0101 1100) XOR (0001 1011) = (0100 0111) Step3: (0101 0111) x (000 1000) = (1000 1110) Step4: (0101 0111) x (0001 000) = (0001 1100) XOR (0001 1011) = (000 0111) Step5: (0101 0111) x (0010 000) = (000 1110) Step6: (0101 0111) x (0100 000) = (0001 1100) Step7: (0101 0111) x (1000 000) = (0011 1000) So, (0101 0111) x (1000 0011) = (0101 0111) x [(000 0001) x (000 0010) x (1000 000)] = (0101 0111) XOR (1010 1110) XOR (0011 1000) = (1100 0001) … [cont.]
Asked by Student1 - Thu Jul 17 17:11:24 2008 - - 1 Answers - 0 Comments

A. Hi, Try these websites: <==Polynomial factoring tool using binary numbers I hope they are of help to you!! :-)
Answered by Pi R Squared - Fri Jul 18 08:10:28 2008

help adding hexadecimal?
Q. 4F1A5 => 0100 111 0001 1010 0101 +B8D5 => 1011 1000 1101 0101 --- I got => ??? 1010 1010 0111 1010 How do you add the 0100 part since bottom is short one number.
Asked by jabba - Wed Feb 4 22:46:59 2009 - - 2 Answers - 0 Comments

A. Just carry the one and bring down the 0101 just like you would adding any other base. You get 0101 1010 1010 0111 1010 = 5 A A 7 A Converting to binary makes a lot of sense to me.
Answered by a simple man - Wed Feb 4 23:45:56 2009

I need an explanation for Echelon form and Reduced form?
Q. | 1 0 2 0 | | 0 1 1 0 | | 0 0 2 0 | | 0 0 0 1 | the matrix above is in echelon form Is it possible to make this matrix in reduced echelon form? When I plug in the martix above on a ti calculator (rref()) i get a reduced echelon form |1000| |0100| |0010| |0001| I need some explanation. Either the calculator is wrong or im forgetting a fundamental rule. Also the case |900| |030| |002| rref() with calculator gives me the identity matrix. why? some explanation would be helpful | 1 0 2 0 | | 0 1 1 0 | | 0 0 2 0 | ~ | 0 0 0 1 | R3 - R1 | 1 0 0 0 | | 0 1 1 0 | | 0 0 2 0 | | 0 0 0 1 | (1/2)R3-R2 | 1 0 0 0 | | 0 1 0 0 | | 0 0 2 0 | <--- | 0 0 0 1 | can I do anything here? comparing the matrix above and below, both have different… [cont.]
Asked by FallOut - Fri Nov 13 13:04:04 2009 - - 1 Answers - 0 Comments

A. Echelon form means that there is a Leading number in each column followed by zeros below it. Also, it needs to be in the order such that the number with the lowest column number has the lowest row number as well. In your example: | 1 0 2 0 | | 0 1 1 0 | | 0 0 2 0 | | 0 0 0 1 | This is in echelon form because under each leading number, there are only zeros. There is a leading number at Row1,Col1 at Row2,Col2 etc. However, the following is also in echelon form by the same rule: | 1 0 2 | | 0 1 1 | | 0 0 0 | Reduced row echelon form means that every column with a Leading One has all zeros in the rest of that column. |1005| |0100| |0010| |0001| This is not in rref form because the 5 is in the same column as the leading 1. rref( |900| |030| |0 [cont.]
Answered by Mark - Fri Nov 13 13:14:29 2009