[tex]\begin{gathered} x^4-x^3-3x^2+x+2 \\ a)\text{ P(2)=0 means that x-2 is a factor of the polynomial} \\ b)\text{ I w}ill\text{ compute the synthetic division below} \end{gathered}[/tex][tex]\begin{gathered} \text{hence,} \\ \frac{x^4-x^3-3x^2+x+2}{x-2}\text{ has quotient equal to } \\ x^3+x^2-x-1 \end{gathered}[/tex][tex]\begin{gathered} c)\text{ In order to solve this point we must factorize the quotient},\text{ i.e.} \\ x^3+x^2-x-1=x^2(x+1)-(x+1) \\ x^3+x^2-x-1=(x+1)(x^2-1) \\ x^3+x^2-x-1=(x+1)(x+1)(x-1) \end{gathered}[/tex][tex]\begin{gathered} d)\text{ hence, we obtain above that:} \\ \frac{x^4-x^3-3x^2+x+2}{x-2}=x^3+x^2-x-1 \\ \text{with the last result, this is equal to:} \\ \frac{x^4-x^3-3x^2+x+2}{x-2}=(x+1)(x+1)(x-1) \\ \text{which implies that} \\ x^4-x^3-3x^2+x+2=(x+1)(x+1)(x-1)(x-2) \\ \text{Therefore, from this last result, the roots of the given poluynomial are:} \\ x=-1,x=-1,x=1\text{ and x=2} \end{gathered}[/tex]
