Create a problem that has a binomial which is not a factor of the polynomial being divided.

5 answers

Consider the polynomial \(x^3 - 3x^2 + 3x - 1\) being divided by the binomial \(x - 2\). In this problem, the binomial \(x - 2\) is not a factor of the polynomial being divided, as there is no integer value of \(x\) for which the polynomial evaluates to zero when divided by \(x - 2\).
create a more understandable equation
Let's consider the polynomial \(2x^3 - 5x^2 + 4x - 1\) being divided by the binomial \(x - 3\). In this case, the binomial \(x - 3\) is not a factor of the polynomial being divided because when we substitute \(x = 3\) into the polynomial, it does not evaluate to zero.
create a problem with a trinomial divided by a binomial and is not a factor
A trinomial being divided by a binomial that is not a factor can be illustrated by considering the polynomial division of \(x^3 + 4x^2 - 5x + 2\) by the binomial \(x - 3\). In this case, the binomial \(x - 3\) is not a factor of the trinomial \(x^3 + 4x^2 - 5x + 2\) as there is no integer value of \(x\) for which the polynomial evaluates to zero when divided by \(x - 3\).