First, we need to find which number when substituted into the equation will give the answer zero. \(f(1) = {(1)^3} + 4{(1)^2} + (1) - 6 = 0\) Therefore \((x - 1)\)is a factor. Factorise the quadratic ...
We show that the binary expansions of algebraic numbers do not form secure pseudorandom sequences; given sufficiently many initial bits of an algebraic number, its minimal polynomial can be ...
Some results have been hidden because they may be inaccessible to you
Show inaccessible results
Feedback