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 ...

Statement and proof of the Factor Theorem. Factorisation of, ax2+bx+c, a is not equal to 0, where a,b, c are real numbers, and of cubic polynomials using the Factor Theorem. Recall of algebraic ...