The FTOA tells us that there are four zeros counting multiplicity. Other methods help us find them: \displaystyle-{2} , \displaystyle-{3} , \displaystyle{2}{i} , \displaystyle-{2}{i} ...

Zeros of \displaystyle{x}^{{4}}−{4}{x}^{{3}}−{35}{x}^{{2}}+{6}{x}+{144} are \displaystyle-{3} , \displaystyle{2} and \displaystyle{8} Explanation: To find all the zeros of \displaystyle{P}{\left({X}\right)}={x}^{{4}}−{4}{x}^{{3}}−{35}{x}^{{2}}+{6}{x}+{144} ...

Use the rational root theorem to help find the first zero, then factor by it and by grouping to find the other roots. Explanation: \displaystyle{h}{\left({x}\right)}={x}^{{4}}+{10}{x}^{{3}}+{26}{x}^{{2}}+{10}{x}+{25} ...

Remainder is \displaystyle{9} Explanation: \displaystyle{x}-{1}={0}\therefore{x}={1} \displaystyle{x}^{{5}}-{2}{x}^{{4}}-{3}{x}^{{3}}+{4}{x}^{{2}}+{3}{x}+{6} By putting \displaystyle{x}={1} ...

The easiest way, but unfortunatelly only works some times is the following: Step 1: Guess a root a of P(x). Step 2: Divide by x-a. Step 3: Factor the remaining quadratic. The following usually ...

By the rational root theorem , f has no rational roots. Since the degree is 3, it follows that f is irreducible. Eisenstein criterion with p=3 is another way to show this.