(New page: Use long division like in Example 1 except the coefficients belong to <math>Z_7</math>. My final answer was <math>q=4x^2+3x+6</math> and <math>r=6x</math>. Can anyone confirm?) |
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Use long division like in Example 1 except the coefficients belong to <math>Z_7</math>. | Use long division like in Example 1 except the coefficients belong to <math>Z_7</math>. | ||
My final answer was <math>q=4x^2+3x+6</math> and <math>r=6x</math>. Can anyone confirm? | My final answer was <math>q=4x^2+3x+6</math> and <math>r=6x</math>. Can anyone confirm? | ||
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| + | I have the same thing for the quotient, but I have 6x+2 for the remainder. | ||
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| + | -Ozgur | ||
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| + | To verify my solution to f/g, where f, g are the given polynomials, I just checked that f = qg + r, where q is my quotient, and r is my remainder. This is how you can be sure. | ||
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| + | I got the same as Ozgur, and I verified it by multiplying and adding the remainder. --[[User:Dakinsey|Dakinsey]] 11:27, 5 November 2008 (UTC) | ||
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| + | I also got a quotient of 4x^2+3x+6 and a remainder of 6x+2..... | ||
Latest revision as of 04:14, 6 November 2008
Use long division like in Example 1 except the coefficients belong to $ Z_7 $. My final answer was $ q=4x^2+3x+6 $ and $ r=6x $. Can anyone confirm?
I have the same thing for the quotient, but I have 6x+2 for the remainder.
-Ozgur
To verify my solution to f/g, where f, g are the given polynomials, I just checked that f = qg + r, where q is my quotient, and r is my remainder. This is how you can be sure.
I got the same as Ozgur, and I verified it by multiplying and adding the remainder. --Dakinsey 11:27, 5 November 2008 (UTC)
I also got a quotient of 4x^2+3x+6 and a remainder of 6x+2.....
