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 ...

Fibonacci cubes represent a fascinating class of graphs that emerge as subgraphs of the n-dimensional hypercube. Defined by the restriction to binary strings that avoid consecutive 1s, these graphs ...

This is a preview. Log in through your library . Abstract For any associative ring $A$ with 1 of prime characteristic $\neq 0, 2, 3$, every element of $A$ is the sum ...

This paper discusses some new integer factoring methods involving cyclotomic polynomials. There are several polynomials $f(X)$ known to have the following property ...