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 ...

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

If \((x \pm h)\) is a factor of a polynomial, then the remainder will be zero. Conversely, if the remainder is zero, then \((x \pm h)\) is a factor. Often ...

This is a preview. Log in through your library . Abstract We show that there exist two cubic polynomials with connected Julia sets which are combinatorially equivalent but not topologically conjugate ...

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 ...

Before being mortally wounded in a duel at age 20, Évariste Galois discovered the hidden structure of polynomial equations. By studying the relationships between their solutions — rather than the ...