Could use expansion (x^5+1)= (x+1)*(x^4-x^3+x^2+-x+1) = (x+1)*q(x). Here numerator, n(x) = (x+1)*q(x)+(x+1) = (x+1)*(q(x)+1). Also denominator, d(x) = (q(x)+1) hence your result n(x)/d(x) = (x+1).
If we let N be 5^75 remain and ignore the remainder of N, and if we let D be 5^60 and ignore the remainder of D, then our answer is 5^75/5^60 = 5^15 which is wrong by 1/5^15 OR too small to care -- I am an engineer first and a mathematician second
Let t = 5^15, and the given E = N/D where N = t^5 +t +2, D = t^4 -t^3 +t^2 -t +2. Then, manipulating N N = t^5 +t^2 -t^2 -t +2t +2; N = (t^2)(t^3 +1) -t(t +1) +2(t+1); N = (t^2)(t+1)(t^2 -t +1) -t(t +1) +2(t+1); N = (t +1)[(t^2)(t^2 -t +1) -t +2]; N = (t +1)[t^4 -t^3 +t^2 -t +2]; that is, N = (t +1)[D] Therefore E = N/D = (t +1)D/D = t +1 E = (5^15) +1
Could use expansion (x^5+1)= (x+1)*(x^4-x^3+x^2+-x+1) = (x+1)*q(x). Here numerator, n(x) = (x+1)*q(x)+(x+1) = (x+1)*(q(x)+1).
Also denominator, d(x) = (q(x)+1) hence your result n(x)/d(x) = (x+1).
Thanks a lot for letting me know various ways of simplification, Sir ^.^
5^15+1
Wonderful introduction clearly explained...
I solved it .thanks
(5^15)+1
If we let N be 5^75 remain and ignore the remainder of N, and if we let D be 5^60 and ignore the remainder of D, then our answer is 5^75/5^60 = 5^15 which is wrong by 1/5^15 OR too small to care -- I am an engineer first and a mathematician second
M. number 1 is excellent.
M. number 2 is easiest.
But in M number 3: what is the basis of choosing (x + 1).
Thanks 🙏.
As method-2 works with division so method-3 should work with multiplication by the same.
Thanks.
Let t = 5^15,
and the given E = N/D
where
N = t^5 +t +2,
D = t^4 -t^3 +t^2 -t +2.
Then, manipulating N
N = t^5 +t^2 -t^2 -t +2t +2;
N = (t^2)(t^3 +1)
-t(t +1) +2(t+1);
N = (t^2)(t+1)(t^2 -t +1)
-t(t +1) +2(t+1);
N = (t +1)[(t^2)(t^2 -t +1)
-t +2];
N = (t +1)[t^4 -t^3 +t^2
-t +2];
that is,
N = (t +1)[D]
Therefore
E = N/D = (t +1)D/D
= t +1
E = (5^15) +1
( x ➖ 3x+2)